The electrostatic potential on the surface of uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of shell, at a distance $r = 5 \ cm$ from centre, and at a distance $r = 15 \ cm$ from the centre of the shell respectively, are:
120V, 120V, 80V
Understanding Electrostatic Potential on a Spherical Shell
For a uniformly charged spherical shell, the electric field and potential behave differently inside and outside the shell. Key properties are:
- Inside the shell ($r < R$): The electric field is zero. Consequently, the electric potential is constant and equal to the potential on the surface.
- Outside the shell ($r > R$): The electric field and potential are the same as if all the charge were concentrated at the center, behaving like a point charge.
Potential Calculations
Given: Radius $R = 10 \ cm$, Surface Potential $V_{surface} = 120 \ V$.
Potential at the Centre ($r = 0$)
The center is inside the shell. Therefore, the potential at the center is equal to the potential on the surface.
$V_{centre} = V_{surface} = 120 \ V$
Potential Inside the Shell ($r = 5 \ cm$)
Since $r = 5 \ cm < R = 10 \ cm$, this point is inside the shell. The potential is constant inside.
$V_{r=5cm} = V_{surface} = 120 \ V$
Potential Outside the Shell ($r = 15 \ cm$)
Since $r = 15 \ cm > R = 10 \ cm$, this point is outside the shell. The potential follows the inverse square law relative to the distance from the center, similar to a point charge.
We know $V_{surface} = \frac{kQ}{R}$, where $kQ$ is the total charge effect. Thus, $kQ = V_{surface} \times R = 120 \ V \times 10 \ cm$.
The potential at a distance $r$ outside the shell is $V_r = \frac{kQ}{r}$.
Substituting the value of $kQ$: $V_{r=15cm} = \frac{(120 \ V \times 10 \ cm)}{15 \ cm}$
Calculating the value: $V_{r=15cm} = 120 \ V \times \frac{10}{15} = 120 \ V \times \frac{2}{3} = 80 \ V$
$V_{r=15cm} = 80 \ V$
Conclusion
The potentials at the centre, at $r = 5 \ cm$, and at $r = 15 \ cm$ are $120 \ V$, $120 \ V$, and $80 \ V$, respectively.
Similar Questions
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
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Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
A wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be
(DROPPED)
A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :
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(Given $\mu_0 = 4\pi \times 10^{-7}$ Tm $A^{-1}$)
