Question

Two points charges 4q and –q are fixed on the x-axis at x = \[ - \left( {\dfrac{d}{2}} \right)\] and x =\[\left( {\dfrac{d}{2}} \right)\] , respectively. If a third point charge ‘q’ is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:

A. decrease by \[\dfrac{q^2}{{4\pi {\varepsilon _0}d}}\]
B. decrease by \[\dfrac{{4q^2}}{{3\pi {\varepsilon _0}d}}\]
C. increase by \[\dfrac{{3q^2}}{{4\pi {\varepsilon _0}d}}\]
D. increase by \[\dfrac{{2{q^2}}}{{3\pi {\varepsilon _0}d}}\]

Answer 1

Hint: To solve this question first we need to find initial potential energy and final potential energy. After subtracting the both we get the change in potential energy. Electric potential energy is defined as the energy that is required to move a charge opposite to the electric field.

Formula used:
The potential energy is given as,
\[U = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{r}\]
Where q is the charge, r is the distance and \[{\varepsilon _0}\] is the permittivity.

Complete step by step solution:

Image: Two point charges 4q and –q are at a distance of \[ - \left( {\dfrac{d}{2}} \right)\] and \[\left( {\dfrac{d}{2}} \right)\] on the x-axis with respect to origin.

As we know the initial potential energy is given as,
\[{U_i} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{4{q^2}}}{{\left( {\dfrac{d}{2}} \right)}} - \dfrac{{{q^2}}}{{\left( {\dfrac{d}{2}} \right)}}} \right]\]
As the final potential energy is given as,
\[{U_f} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{4{q^2}}}{{\left( {\dfrac{{3d}}{2}} \right)}} - \dfrac{{{q^2}}}{{\left( {\dfrac{d}{2}} \right)}}} \right]\]

Now the change in potential energy will be
\[\Delta U = {U_f} - {U_i}\]
\[\Rightarrow \Delta U =\dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{4{q^2}}}{{\left( {\dfrac{{3d}}{2}} \right)}} - \dfrac{{{q^2}}}{{\left( {\dfrac{d}{2}} \right)}}} \right]\]-\[\dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{4{q^2}}}{{\left( {\dfrac{d}{2}} \right)}} - \dfrac{{{q^2}}}{{\left( {\dfrac{d}{2}} \right)}}} \right]\]
\[\therefore \Delta U = - \dfrac{{4{q^2}}}{{3\pi {\varepsilon _0}d}}\]
Here the negative sign shows the decreasing energy. Therefore, the energy of the charge will decrease by \[\dfrac{{4{q^2}}}{{3\pi {\varepsilon _0}d}}\].

Hence option B is the correct answer.

Note: As we know the energy that is needed to move a charge against an electric field. We need greater energy to accelerate a charge in the electric field but also greater energy to move it through a stronger electric field. At any given point in an electric field the electric potential energy is the electric potential multiplied by the amount of charge at that point. So, the voltage gives you how much potential energy could have at some point which is not dependent on how much charge you put there.