How to Solve Laplace's Equation in Spherical Coordinates

General Solution

Use the ansatz V ( r , θ ) = R ( r ) Θ ( θ ) {\displaystyle V(r,\theta )=R(r)\Theta (\theta )} V(r,\theta )=R(r)\Theta (\theta ) and substitute it into the equation. In the most general case, the potential depends on all three variables. However, in many physical scenarios, there exists an azimuthal symmetry to the problem. For a physical example, an insulating sphere could have a charge density that is only dependent on θ , {\displaystyle \theta ,} \theta , so the potential must not depend on ϕ . {\displaystyle \phi .} \phi . This assumption greatly simplifies the problem so that we do not have to deal with spherical harmonics. First, we simply substitute. Θ r 2 ∂ ∂ r ( r 2 ∂ R ∂ r ) + R r 2 sin ⁡ θ ∂ ∂ θ ( sin ⁡ θ ∂ Θ ∂ θ ) = 0 {\displaystyle {\frac {\Theta }{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial R}{\partial r}}\right)+{\frac {R}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial \Theta }{\partial \theta }}\right)=0} {\frac {\Theta }{r^{{2}}}}{\frac {\partial }{\partial r}}\left(r^{{2}}{\frac {\partial R}{\partial r}}\right)+{\frac {R}{r^{{2}}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial \Theta }{\partial \theta }}\right)=0 Divide the equation by V . {\displaystyle V.} V. What remains is a term that only depends on r {\displaystyle r} r and a term that only depends on θ . {\displaystyle \theta .} \theta . The derivatives then become ordinary derivatives. 1 R d d r ( r 2 d R d r ) + 1 Θ sin ⁡ θ d d θ ( sin ⁡ θ d Θ d θ ) = 0 {\displaystyle {\frac {1}{R}}{\frac {\mathrm {d} }{\mathrm {d} r}}\left(r^{2}{\frac {\mathrm {d} R}{\mathrm {d} r}}\right)+{\frac {1}{\Theta \sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)=0} {\frac {1}{R}}{\frac {{\mathrm {d}}}{{\mathrm {d}}r}}\left(r^{{2}}{\frac {{\mathrm {d}}R}{{\mathrm {d}}r}}\right)+{\frac {1}{\Theta \sin \theta }}{\frac {{\mathrm {d}}}{{\mathrm {d}}\theta }}\left(\sin \theta {\frac {{\mathrm {d}}\Theta }{{\mathrm {d}}\theta }}\right)=0

Set the two terms equal to constants. An argument must be made here. We have a term that only depends on r {\displaystyle r} r and a term that only depends on θ . {\displaystyle \theta .} \theta . Their sum, however, must always equal 0. Since these derivatives are varying quantities in general, the only way that this can be true for all values of r {\displaystyle r} r and θ {\displaystyle \theta } \theta is if the terms are both constant. We will see very shortly that it is convenient for us to denote the constant by l ( l + 1 ) . {\displaystyle l(l+1).} l(l+1). 1 R d d r ( r 2 d R d r ) = l ( l + 1 ) {\displaystyle {\frac {1}{R}}{\frac {\mathrm {d} }{\mathrm {d} r}}\left(r^{2}{\frac {\mathrm {d} R}{\mathrm {d} r}}\right)=l(l+1)} {\frac {1}{R}}{\frac {{\mathrm {d}}}{{\mathrm {d}}r}}\left(r^{{2}}{\frac {{\mathrm {d}}R}{{\mathrm {d}}r}}\right)=l(l+1) 1 Θ sin ⁡ θ d d θ ( sin ⁡ θ d Θ d θ ) = − l ( l + 1 ) {\displaystyle {\frac {1}{\Theta \sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)=-l(l+1)} {\frac {1}{\Theta \sin \theta }}{\frac {{\mathrm {d}}}{{\mathrm {d}}\theta }}\left(\sin \theta {\frac {{\mathrm {d}}\Theta }{{\mathrm {d}}\theta }}\right)=-l(l+1) We have now converted Laplace's equation, assuming azimuthal symmetry, into two noncoupled ordinary differential equations.

Solve the radial equation. After multiplying and using the product rule, we find that this is simply the Euler-Cauchy equation. r 2 d 2 R d r 2 + 2 r d R d r − l ( l + 1 ) R = 0 {\displaystyle r^{2}{\frac {\mathrm {d} ^{2}R}{\mathrm {d} r^{2}}}+2r{\frac {\mathrm {d} R}{\mathrm {d} r}}-l(l+1)R=0} r^{{2}}{\frac {{\mathrm {d}}^{{2}}R}{{\mathrm {d}}r^{{2}}}}+2r{\frac {{\mathrm {d}}R}{{\mathrm {d}}r}}-l(l+1)R=0 The standard method of solving this equation is to assume the solution of the form R = r n {\displaystyle R=r^{n}} R=r^{{n}} and solve the resulting characteristic equation. In particular, we expand the quantity in the square root and factor. n 2 + n − l ( l + 1 ) = 0 {\displaystyle n^{2}+n-l(l+1)=0} n^{{2}}+n-l(l+1)=0 n = − 1 2 ± 1 + 4 l ( l + 1 ) 2 = l , − l − 1 {\displaystyle n=-{\frac {1}{2}}\pm {\frac {\sqrt {1+4l(l+1)}}{2}}=l,\ -l-1} n=-{\frac {1}{2}}\pm {\frac {{\sqrt {1+4l(l+1)}}}{2}}=l,\ -l-1 The roots of the characteristic equation suggest our choice of constant. Since the Euler-Cauchy equation is a linear equation, the solution to the radial part is as follows. R ( r ) = A r l + B r l + 1 {\displaystyle R(r)=Ar^{l}+{\frac {B}{r^{l+1}}}} R(r)=Ar^{{l}}+{\frac {B}{r^{{l+1}}}}

Solve the angular equation. This equation is the Legendre differential equation in the variable cos ⁡ θ . {\displaystyle \cos \theta .} \cos \theta . 1 sin ⁡ θ d d θ ( sin ⁡ θ d Θ d θ ) + l ( l + 1 ) Θ = 0 {\displaystyle {\frac {1}{\sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)+l(l+1)\Theta =0} {\frac {1}{\sin \theta }}{\frac {{\mathrm {d}}}{{\mathrm {d}}\theta }}\left(\sin \theta {\frac {{\mathrm {d}}\Theta }{{\mathrm {d}}\theta }}\right)+l(l+1)\Theta =0 To see this, we begin with the Legendre equation in the variable x {\displaystyle x} x and make the substitution x = cos ⁡ θ , {\displaystyle x=\cos \theta ,} x=\cos \theta , implying that d x = − sin ⁡ θ d θ . {\displaystyle \mathrm {d} x=-\sin \theta \mathrm {d} \theta .} {\mathrm {d}}x=-\sin \theta {\mathrm {d}}\theta . d d x ( ( 1 − x 2 ) d Θ d x ) + l ( l + 1 ) Θ = 0 d − sin ⁡ θ d θ ( sin 2 ⁡ θ d Θ − sin ⁡ θ d θ ) + l ( l + 1 ) Θ = 0 1 sin ⁡ θ d d θ ( sin ⁡ θ d Θ d θ ) + l ( l + 1 ) Θ = 0 {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\left((1-x^{2}){\frac {\mathrm {d} \Theta }{\mathrm {d} x}}\right)+l(l+1)\Theta &=0\\{\frac {\mathrm {d} }{-\sin \theta \mathrm {d} \theta }}\left(\sin ^{2}\theta {\frac {\mathrm {d} \Theta }{-\sin \theta \mathrm {d} \theta }}\right)+l(l+1)\Theta &=0\\{\frac {1}{\sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)+l(l+1)\Theta &=0\end{aligned}}} {\begin{aligned}{\frac {{\mathrm {d}}}{{\mathrm {d}}x}}\left((1-x^{{2}}){\frac {{\mathrm {d}}\Theta }{{\mathrm {d}}x}}\right)+l(l+1)\Theta &=0\\{\frac {{\mathrm {d}}}{-\sin \theta {\mathrm {d}}\theta }}\left(\sin ^{{2}}\theta {\frac {{\mathrm {d}}\Theta }{-\sin \theta {\mathrm {d}}\theta }}\right)+l(l+1)\Theta &=0\\{\frac {1}{\sin \theta }}{\frac {{\mathrm {d}}}{{\mathrm {d}}\theta }}\left(\sin \theta {\frac {{\mathrm {d}}\Theta }{{\mathrm {d}}\theta }}\right)+l(l+1)\Theta &=0\end{aligned}} This equation can be solved using the method of Frobenius. In particular, the solutions are Legendre polynomials in cos ⁡ θ , {\displaystyle \cos \theta ,} \cos \theta , which we write as P l ( cos ⁡ θ ) . {\displaystyle P_{l}(\cos \theta ).} P_{{l}}(\cos \theta ). These are orthogonal polynomials with respect to an inner product, which we elaborate on shortly. This orthogonality means that we can write any polynomial as a linear combination of Legendre polynomials. Θ ( θ ) = P l ( cos ⁡ θ ) {\displaystyle \Theta (\theta )=P_{l}(\cos \theta )} \Theta (\theta )=P_{{l}}(\cos \theta ) The first few Legendre polynomials are given as follows. Notice that the polynomials alternate between even and odd. These polynomials will be very important in the next sections. P 0 ( x ) = 1 {\displaystyle P_{0}(x)=1} P_{{0}}(x)=1 P 1 ( x ) = x {\displaystyle P_{1}(x)=x} P_{{1}}(x)=x P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) {\displaystyle P_{2}(x)={\frac {1}{2}}(3x^{2}-1)} P_{{2}}(x)={\frac {1}{2}}(3x^{{2}}-1) P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) {\displaystyle P_{3}(x)={\frac {1}{2}}(5x^{3}-3x)} P_{{3}}(x)={\frac {1}{2}}(5x^{{3}}-3x) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) {\displaystyle P_{4}(x)={\frac {1}{8}}(35x^{4}-30x^{2}+3)} P_{{4}}(x)={\frac {1}{8}}(35x^{{4}}-30x^{{2}}+3) It turns out that there is another solution to the Legendre differential equation. However, this solution cannot be part of the general solution because it blows up at θ = 0 {\displaystyle \theta =0} \theta =0 and θ = π , {\displaystyle \theta =\pi ,} \theta =\pi , so it is omitted. Θ ( θ ) = ln ⁡ tan ⁡ θ 2 {\displaystyle \Theta (\theta )=\ln \tan {\frac {\theta }{2}}} \Theta (\theta )=\ln \tan {\frac {\theta }{2}}

Construct the general solution. We now have our solutions to both the radial and angular equations. We can then write out the general solution as a series, since by linearity, any linear combination of these solutions is also a solution. V ( r , θ ) = ∑ l = 0 ∞ ( A l r l + B l r l + 1 ) P l ( cos ⁡ θ ) {\displaystyle V(r,\theta )=\sum _{l=0}^{\infty }\left(A_{l}r^{l}+{\frac {B_{l}}{r^{l+1}}}\right)P_{l}(\cos \theta )} V(r,\theta )=\sum _{{l=0}}^{{\infty }}\left(A_{{l}}r^{{l}}+{\frac {B_{{l}}}{r^{{l+1}}}}\right)P_{{l}}(\cos \theta )

Boundary Conditions

Assume that a sphere with radius R {\displaystyle R} R contains a potential on its surface. This is an example of a Dirichlet boundary condition, where the value everywhere on the boundary is specified. We then proceed to solve for the coefficients A l {\displaystyle A_{l}} A_{{l}} and B l . {\displaystyle B_{l}.} B_{{l}}. V ( R , θ ) = V 0 ( θ ) {\displaystyle V(R,\theta )=V_{0}(\theta )} V(R,\theta )=V_{{0}}(\theta ) V 0 ( θ ) = ∑ l = 0 ∞ ( A l R l + B l R l + 1 ) P l ( cos ⁡ θ ) {\displaystyle V_{0}(\theta )=\sum _{l=0}^{\infty }\left(A_{l}R^{l}+{\frac {B_{l}}{R^{l+1}}}\right)P_{l}(\cos \theta )} V_{{0}}(\theta )=\sum _{{l=0}}^{{\infty }}\left(A_{{l}}R^{{l}}+{\frac {B_{{l}}}{R^{{l+1}}}}\right)P_{{l}}(\cos \theta )

Find the potential inside the sphere. Physically, the potential cannot blow up at the origin, so B l = 0 {\displaystyle B_{l}=0} B_{{l}}=0 for all l . {\displaystyle l.} l. Multiply both sides by P l ′ ( cos ⁡ θ ) sin ⁡ θ {\displaystyle P_{l'}(\cos \theta )\sin \theta } P_{{l'}}(\cos \theta )\sin \theta and integrate from 0 {\displaystyle 0} {\displaystyle 0} to π {\displaystyle \pi } \pi . The Legendre polynomials are orthogonal with respect to this inner product. ∫ 0 π V 0 ( θ ) P l ′ ( cos ⁡ θ ) sin ⁡ θ d θ = ∑ l = 0 ∞ A l R l ∫ 0 π P l ( cos ⁡ θ ) P l ′ ( cos ⁡ θ ) sin ⁡ θ d θ {\displaystyle \int _{0}^{\pi }V_{0}(\theta )P_{l'}(\cos \theta )\sin \theta \mathrm {d} \theta =\sum _{l=0}^{\infty }A_{l}R^{l}\int _{0}^{\pi }P_{l}(\cos \theta )P_{l'}(\cos \theta )\sin \theta \mathrm {d} \theta } \int _{{0}}^{{\pi }}V_{{0}}(\theta )P_{{l'}}(\cos \theta )\sin \theta {\mathrm {d}}\theta =\sum _{{l=0}}^{{\infty }}A_{{l}}R^{{l}}\int _{{0}}^{{\pi }}P_{{l}}(\cos \theta )P_{{l'}}(\cos \theta )\sin \theta {\mathrm {d}}\theta We take advantage of the very important relation, written below. δ l l ′ {\displaystyle \delta _{ll'}} \delta _{{ll'}} is the Kronecker delta, meaning that the integral is nonzero only when l = l ′ . {\displaystyle l=l'.} l=l'. ∫ 0 π P l ( cos ⁡ θ ) P l ′ ( cos ⁡ θ ) sin ⁡ θ d θ = 2 2 l + 1 δ l l ′ {\displaystyle \int _{0}^{\pi }P_{l}(\cos \theta )P_{l'}(\cos \theta )\sin \theta \mathrm {d} \theta ={\frac {2}{2l+1}}\delta _{ll'}} \int _{{0}}^{{\pi }}P_{{l}}(\cos \theta )P_{{l'}}(\cos \theta )\sin \theta {\mathrm {d}}\theta ={\frac {2}{2l+1}}\delta _{{ll'}} ∫ 0 π V 0 ( θ ) P l ( cos ⁡ θ ) sin ⁡ θ d θ = ∑ l = 0 ∞ A l R l 2 2 l + 1 {\displaystyle \int _{0}^{\pi }V_{0}(\theta )P_{l}(\cos \theta )\sin \theta \mathrm {d} \theta =\sum _{l=0}^{\infty }A_{l}R^{l}{\frac {2}{2l+1}}} \int _{{0}}^{{\pi }}V_{{0}}(\theta )P_{{l}}(\cos \theta )\sin \theta {\mathrm {d}}\theta =\sum _{{l=0}}^{{\infty }}A_{{l}}R^{{l}}{\frac {2}{2l+1}}

Solve for A l {\displaystyle A_{l}} A_{{l}}. Knowing the coefficients, we have our potential inside the sphere in terms of a series, with the coefficients written in terms of integrals that, in principle, can be calculated. Note that this method only works because the Legendre polynomials constitute a complete set on the interval 0 ≤ θ ≤ π . {\displaystyle 0\leq \theta \leq \pi .} 0\leq \theta \leq \pi . A l = 2 l + 1 2 R l ∫ 0 π V 0 ( θ ) P l ( cos ⁡ θ ) sin ⁡ θ d θ {\displaystyle A_{l}={\frac {2l+1}{2R^{l}}}\int _{0}^{\pi }V_{0}(\theta )P_{l}(\cos \theta )\sin \theta \mathrm {d} \theta } A_{{l}}={\frac {2l+1}{2R^{{l}}}}\int _{{0}}^{{\pi }}V_{{0}}(\theta )P_{{l}}(\cos \theta )\sin \theta {\mathrm {d}}\theta

Find the potential outside the sphere. We typically set the potential to 0 at infinity. This means that A l = 0. {\displaystyle A_{l}=0.} A_{{l}}=0. Using the same method, we can find the coefficients of B l . {\displaystyle B_{l}.} B_{{l}}. B l = 2 l + 1 2 R l + 1 ∫ 0 π V 0 ( θ ) P l ( cos ⁡ θ ) sin ⁡ θ d θ {\displaystyle B_{l}={\frac {2l+1}{2}}R^{l+1}\int _{0}^{\pi }V_{0}(\theta )P_{l}(\cos \theta )\sin \theta \mathrm {d} \theta } B_{{l}}={\frac {2l+1}{2}}R^{{l+1}}\int _{{0}}^{{\pi }}V_{{0}}(\theta )P_{{l}}(\cos \theta )\sin \theta {\mathrm {d}}\theta

Electric Potential

Find the electric potential everywhere, given a potential on the surface of a sphere of radius R {\displaystyle R} R. The surface has a potential V 0 ( θ ) = k cos ⁡ 4 θ , {\displaystyle V_{0}(\theta )=k\cos 4\theta ,} V_{{0}}(\theta )=k\cos 4\theta , where k {\displaystyle k} k is a constant. The goal of problems like these is to solve for the coefficients A l {\displaystyle A_{l}} A_{{l}} and B l . {\displaystyle B_{l}.} B_{{l}}. From the previous section, we could in principle just do the integrals...but we opt to save some labor by comparing coefficients. V ( R , θ ) = V 0 ( θ ) = ∑ l = 0 ∞ ( A l R l + B l R l + 1 ) P l ( cos ⁡ θ ) {\displaystyle V(R,\theta )=V_{0}(\theta )=\sum _{l=0}^{\infty }\left(A_{l}R^{l}+{\frac {B_{l}}{R^{l+1}}}\right)P_{l}(\cos \theta )} V(R,\theta )=V_{{0}}(\theta )=\sum _{{l=0}}^{{\infty }}\left(A_{{l}}R^{{l}}+{\frac {B_{{l}}}{R^{{l+1}}}}\right)P_{{l}}(\cos \theta )

Write the potential on the surface in terms of Legendre polynomials. This step is crucial in comparing coefficients, and we can use trigonometric identities to do this. We then refer to the zeroth, second, and fourth polynomials to write V 0 {\displaystyle V_{0}} V_{{0}} in terms of them. k cos ⁡ 4 θ = k ( 8 cos 4 ⁡ θ − 8 cos 2 ⁡ θ + 1 ) = k ( 64 35 P 4 ( cos ⁡ θ ) − 16 21 P 2 ( cos ⁡ θ ) − 1 15 P 0 ( cos ⁡ θ ) ) {\displaystyle {\begin{aligned}k\cos 4\theta &=k(8\cos ^{4}\theta -8\cos ^{2}\theta +1)\\&=k\left({\frac {64}{35}}P_{4}(\cos \theta )-{\frac {16}{21}}P_{2}(\cos \theta )-{\frac {1}{15}}P_{0}(\cos \theta )\right)\end{aligned}}} {\begin{aligned}k\cos 4\theta &=k(8\cos ^{{4}}\theta -8\cos ^{{2}}\theta +1)\\&=k\left({\frac {64}{35}}P_{{4}}(\cos \theta )-{\frac {16}{21}}P_{{2}}(\cos \theta )-{\frac {1}{15}}P_{{0}}(\cos \theta )\right)\end{aligned}}

Solve for the potential outside the sphere. Physically, the potential should go to 0 as r → ∞ . {\displaystyle r\to \infty .} r\to \infty . This means that outside the sphere, A l = 0. {\displaystyle A_{l}=0.} A_{{l}}=0. V ( r ≥ R , θ ) = ∑ l = 0 ∞ B l r l + 1 P l ( cos ⁡ θ ) {\displaystyle V(r\geq R,\theta )=\sum _{l=0}^{\infty }{\frac {B_{l}}{r^{l+1}}}P_{l}(\cos \theta )} V(r\geq R,\theta )=\sum _{{l=0}}^{{\infty }}{\frac {B_{{l}}}{r^{{l+1}}}}P_{{l}}(\cos \theta ) We then compare coefficients (there are three of them) to match boundary conditions. B 0 R = − k 15 , B 0 = − k 15 R {\displaystyle {\frac {B_{0}}{R}}=-{\frac {k}{15}},\ B_{0}=-{\frac {k}{15}}R} {\frac {B_{{0}}}{R}}=-{\frac {k}{15}},\ B_{{0}}=-{\frac {k}{15}}R B 2 R 3 = − 16 k 21 , B 2 = − 16 k 21 R 3 {\displaystyle {\frac {B_{2}}{R^{3}}}=-{\frac {16k}{21}},\ B_{2}=-{\frac {16k}{21}}R^{3}} {\frac {B_{{2}}}{R^{{3}}}}=-{\frac {16k}{21}},\ B_{{2}}=-{\frac {16k}{21}}R^{{3}} B 4 R 5 = 64 k 35 , B 4 = 64 k 35 R 5 {\displaystyle {\frac {B_{4}}{R^{5}}}={\frac {64k}{35}},\ B_{4}={\frac {64k}{35}}R^{5}} {\frac {B_{{4}}}{R^{{5}}}}={\frac {64k}{35}},\ B_{{4}}={\frac {64k}{35}}R^{{5}} Plugging back into the solution, we have the potential outside the sphere. V ( r ≥ R , θ ) = k ( − R 15 r − 16 R 3 21 r 3 P 2 ( cos ⁡ θ ) + 64 R 5 35 r 5 P 4 ( cos ⁡ θ ) ) {\displaystyle V(r\geq R,\theta )=k\left(-{\frac {R}{15r}}-{\frac {16R^{3}}{21r^{3}}}P_{2}(\cos \theta )+{\frac {64R^{5}}{35r^{5}}}P_{4}(\cos \theta )\right)} V(r\geq R,\theta )=k\left(-{\frac {R}{15r}}-{\frac {16R^{{3}}}{21r^{{3}}}}P_{{2}}(\cos \theta )+{\frac {64R^{{5}}}{35r^{{5}}}}P_{{4}}(\cos \theta )\right)

Solve for the potential inside the sphere. Since there is no charge density inside the sphere, the potential cannot blow up, so B l = 0. {\displaystyle B_{l}=0.} B_{{l}}=0. Furthermore, the boundary conditions and this technique ensure that the potential is continuous - in other words, the potential infinitesimally near the surface is the same when approached from both outside and inside the sphere. V ( r ≤ R , θ ) = ∑ l = 0 ∞ A l r l P l ( cos ⁡ θ ) {\displaystyle V(r\leq R,\theta )=\sum _{l=0}^{\infty }A_{l}r^{l}P_{l}(\cos \theta )} V(r\leq R,\theta )=\sum _{{l=0}}^{{\infty }}A_{{l}}r^{{l}}P_{{l}}(\cos \theta ) Again, we compare coefficients to match boundary conditions. A 0 = − k 15 {\displaystyle A_{0}=-{\frac {k}{15}}} A_{{0}}=-{\frac {k}{15}} A 2 = − 16 k 21 R 2 {\displaystyle A_{2}=-{\frac {16k}{21R^{2}}}} A_{{2}}=-{\frac {16k}{21R^{{2}}}} A 4 = 64 k 35 R 4 {\displaystyle A_{4}={\frac {64k}{35R^{4}}}} A_{{4}}={\frac {64k}{35R^{{4}}}} We now have the potential inside the sphere. V ( r ≤ R , θ ) = k ( − 1 15 − 16 r 2 21 R 2 P 2 ( cos ⁡ θ ) + 64 r 4 35 R 4 P 4 ( cos ⁡ θ ) ) {\displaystyle V(r\leq R,\theta )=k\left(-{\frac {1}{15}}-{\frac {16r^{2}}{21R^{2}}}P_{2}(\cos \theta )+{\frac {64r^{4}}{35R^{4}}}P_{4}(\cos \theta )\right)} V(r\leq R,\theta )=k\left(-{\frac {1}{15}}-{\frac {16r^{{2}}}{21R^{{2}}}}P_{{2}}(\cos \theta )+{\frac {64r^{{4}}}{35R^{{4}}}}P_{{4}}(\cos \theta )\right) We can substitute r = R {\displaystyle r=R} r=R in both equations to check for equality. As mentioned before, the potential must be continuous.