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If two charged spheres of radii $R_1$ and $R_2$ have equal surface charge densities, then what will be the ratio of their potential ?
A) $\dfrac{{{R_2}}}{{{R_1}}}$
B) ${(\dfrac{{{R_2}}}{{{R_1}}})^2}$
C) $\dfrac{{{R_1}}}{{{R_2}}}$
D) ${(\dfrac{{{R_1}}}{{{R_2}}})^2}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: An electric potential is the amount of work needed to move a unit charge from infinity to a specific point.
$V = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}$ (q is the charge, r is the distance between infinity and the specific point)
Using the above relation of electric potential we will calculate the ratio of two radii.

Complete step by step solution:
Let's define electric potential and surface charge density in detail.
An electric potential is the amount of work needed to be done to move an electric charge from a reference point to a specific point in an electric field without producing an acceleration. Generally we assume a reference point as earth or infinity. The SI unit of electric potential is volt.
Formula for electric potential is given by:
$V = \dfrac{q}{{4\pi {\varepsilon _0}r}}$
Surface charge density: Surface charge density is the measure of how much electric charge is accumulated over a surface. It is calculated as the charge per unit area. Then the surface charge density is given by:
$\sigma = \dfrac{q}{A}$ ($\sigma $ is the surface charge density, q is the charge and A is the surface area)
Now, we will calculate the ratio of radii of the two spheres.
As we are being given in the question, that the densities of the two sphere are equal then we can have;
$ \Rightarrow \dfrac{{{q_1}}}{{4\pi {\varepsilon _0}{R_1}^2}} = \dfrac{{{q_2}}}{{4\pi {\varepsilon _0}{R_2}^2}}$
On cancelling the common terms, we have
$ \Rightarrow \dfrac{{{q_1}}}{{{q_2}}} = \dfrac{{{R_1}^2}}{{{R_2}^2}}$..........................1
Electric potential is given by:
$ \Rightarrow {V_1} = \dfrac{{{q_1}}}{{4\pi {\varepsilon _0}{R_1}}}$ (potential of sphere 1)...............2
$ \Rightarrow {V_2} = \dfrac{{{q_2}}}{{4\pi {\varepsilon _0}{R_2}}}$ (potential of sphere 2)................3
On dividing equation 2 by 3
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{{{q_1}}}{{4\pi {\varepsilon _0}{R_1}}}}}{{\dfrac{{{q_2}}}{{4\pi {\varepsilon _0}{R_2}}}}}$ (On cancelling the common term and rearranging the fraction)
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{{{q_1}}}{{{R_1}}}}}{{\dfrac{{{q_2}}}{{{R_2}}}}}$(fraction will become)
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{q_1}}}{{{R_1}}} \times \dfrac{{{R_2}}}{{{q_2}}}$(rearranged fraction).............4
On substituting the value of equation 1 in 4
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{R_2}}}{{{R_1}}} \times {(\dfrac{{{R_1}}}{{{R_2}}})^2}$(We will cancel the common term)
$ \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{R_1}}}{{{R_2}}}$

Hence, Option C is correct.

Note: Electric potential is the most fundamental to the storage and release of electrical energy. Electric potential is the common base for the generation of electric fields. Electron microscope works on the principle of high electric potential difference to accelerate electrons in a beam that bombards the sample under examination.