Figure shows a annular disc of outer radius R inner radius R, and uniform surface charge density. Total charge is q. The potential at the centre is (k q/r')(k = (1/4 π (ϵ )0)) where r' is

Question

Figure shows a annular disc of outer radius R inner radius R, and uniform surface charge density. Total charge is q. The potential at the centre is (k q/r')(k = (1/4 π (ϵ )0)) where r' is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-mxpucrujdfbjkmrq.jpg" /> (A) (R + r/2) (B) (R - r/2) (C) R+r (D) R-r

Tardigrade Admin · Accepted Answer

V=k displaystyle ∫ x = rR(σ ⋅ 2 π x d x/x)=k⋅ σ 2π (.R-r.)

Q. Figure shows a annular disc of outer radius $R$ inner radius $R,$ and uniform surface charge density. Total charge is $q .$ The potential at the centre is $\frac{k q}{r ^{^{'}}} (k = \frac{1}{4 π ( ϵ ) _{0}})$ where $r^{^{'}}$ is

$\frac{R + r}{2}$

$\frac{R - r}{2}$

$R + r$

$R - r$

$V = k \int_{x = r}^{R} \frac{σ \cdot 2 π x d x}{x} = k \cdot σ 2 π (R - r)$

Q. Figure shows a annular disc of outer radius R inner radius R, and uniform surface charge density. Total charge is q. The potential at the centre is r′kq​(k=4π(ϵ)0​1​) where r′ is

2R+r​

2R−r​

R+r

R−r