Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

Spherical symmetric charge system centered at origin. 

Electric potential $\phi=\frac{Q}{4 \pi \varepsilon_0 R_0} \quad r \leq R_0$

                         $\phi=\frac{Q}{4 \pi \varepsilon_0 r} \quad r>R_0$

Options:

A spherical symmetry at r = 2R0 encloses a net charge Q

Electric field is discontinued at r = R0

Change is only present at r = R0

All of the above

Correct Answer:

All of the above

Explanation:

For $r>R_0, E=-\frac{d \phi}{d r}=\frac{Q}{4 \pi \varepsilon_0 r^2}$

Charge enclosed by the concentric spherical surface of $r=2 R_0$

$=\varepsilon_0 \phi_{C}=\varepsilon_0 E \times 4 \pi \times r^2=\varepsilon_0 \frac{Q}{4 \pi \varepsilon_0 r^2} 4 \pi r^2=Q$

For $r<R_0, E=-\frac{d \phi}{dr}=0$ (∵ c = constant)

For $r>R_0, E=-\frac{d \phi}{d t}=\frac{Q}{4 \pi \varepsilon_0 r^2}$

Since for $r<R_0, E=0$, hence charge will be only on the spherical surface of $r=R_0$.

For $r<R_0, E=0 \Rightarrow$ energy density $\frac{1}{2} \varepsilon_0 E^2=0$

(d)