Question

A uniformly charged insulating disc of radius $40cm$ has charge per unit area $4.0\,nC/{m^2}$. The electric potential on the axis of the disc at a distance of $30cm$ from its centre will be nearly:
A. $300V$
B. $30V$
C. $450V$
D. $45V$

Answer 1

Hint:First, assume the electric potential at distance be $x$. In the question, the radius of an insulating disc is $40cm$ that has charge per unit area $4.0\,nC/{m^2}$ and $30cm$ from its centre are given. To find the value of electric potential at distance, we need to use the given relation between electric potential at distance and distance from centre to obtain the magnitude.

Formula used:
Electric potential due to a point charge is given by:
$V = \dfrac{{kQ}}{r}$
Where $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ and $r$ is the distance of the point from the centre of the sphere.

Complete step by step solution:
The electric field generated by a point charge is the same for a sphere that is uniformly charged. This indicates that the potential outside the sphere is the same as the potential from a point charge.


Let us assume the electric potential at distance be $x$ from the centre of the sphere. First, convert the radius into its SI unit, then we have $r = 40cm = 40 \times {10^{ - 2}}m$. Similarly the charge per unit area is $Q = 4.0\,nC/{m^2} = 4.0 \times {10^{ - 9}}C/{m^2}$. As we know that $dq = Q \cdot dA$. Here, $dA = 2\pi r \cdot dr$ then we have:
$dq = Q\,2\pi \cdot dr$
Electric potential at distance $x$ from centre is given by:
$V = \dfrac{{kQ}}{r}$
Substitute the given information in the above formula, then:
$dx = k \cdot \dfrac{{dq}}{{\sqrt {{r^2} + {d^2}} }} \\$
$\Rightarrow dx = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{Q\,2\pi r\,dr}}{{\sqrt {{r^2} + {d^2}} }} \\$
Integrate the equation from $0$ to $R$, to determine the total electric potential at the centre point,
$\int {dx} = \dfrac{Q}{{4\pi {\varepsilon _0}}} \cdot \int\limits_0^R {\dfrac{{\,2\pi r\,dr}}{{\sqrt {{r^2} + {d^2}} }}} $
Apply limits on the above equation, then we obtain:
$x = \dfrac{Q}{{2{\varepsilon _0}}} \cdot \{ \sqrt {{r^2} + {d^2}} - x\} $

As we know that ${\varepsilon _0}$ is the permittivity of free space, hence the value of ${\varepsilon _0}$ is $8.85 \times {10^{ - 12}}$. Substitute the given values in the above equation:
$x = \dfrac{{4 \times {{10}^{ - 9}}}}{{2 \times 8.85 \times {{10}^{ - 12}}}}\,(\sqrt {{{(40)}^2} + {{(30)}^2}}) \\$
$\Rightarrow x = \dfrac{{4 \times {{10}^{ - 9}} \times {{10}^{ - 2}}\{ 50 - 30\} }}{{2 \times 8.85 \times {{10}^{ - 12}}}} \\$
$\Rightarrow x = \dfrac{{400}}{{8.85}} \\$
$\therefore x = 45.19\,V $

Therefore, the correct option is D.

Note: Because the relationship between the value of the electric potential at a distance of $30cm$ from its centre and the question can occasionally mislead pupils, read the question carefully. For a location inside the sphere with uniform charging, electric potential is $V = \dfrac{{kQ}}{{{R^3}}}{r^2}$ where $r$ is the distance of point from centre and $R$ is the radius of the sphere. Alternatively, by applying the inverse proportionality of electric potential on distance for the same charge, $V \propto \dfrac{1}{r}$ . This problem can also be resolved without assuming any charge on the sphere.