Question
Two equal point charges are fixed at $$x = - a$$ and $$x = + a$$ on the $$x$$-axis. Another point charge $$Q$$ is placed at the origin. The change in the electrical potential energy of $$Q,$$ when it is displaced by a small distance $$x$$ along the $$x$$-axis, is approximately proportional to
A.
$$x$$
B.
$${x^2}$$
C.
$${x^3}$$
D.
$$\frac{1}{x}$$
Answer :
$${x^2}$$
Solution :
Initial energy
$$\eqalign{ & {U_f} = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{a + x}} + \frac{1}{{a - x}}} \right] = \frac{{2Qqa}}{{4\pi {\varepsilon _0}\left( {{a^2} - {x^2}} \right)}} \cr & {U_i} - {U_f} = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{a} - \frac{a}{{\left( {{a^2} - {x^2}} \right)}}} \right] \cr & = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{a^2} - {x^2} - {a^2}}}{{a\left( {{a^2} - {x^2}} \right)}}} \right] = \frac{{ - 2Qq{x^2}}}{{4\pi {\varepsilon _0}{a^3}}} \cr} $$
when $$x < < a$$ then $${x^2}$$ is neglected in denominator.
$${U_i} - {U_f} = \left( {\frac{{ - Qq}}{{2\pi {\varepsilon _0}{a^3}}}} \right){x^2}$$
Initial energy
$$\eqalign{ & {U_f} = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{a + x}} + \frac{1}{{a - x}}} \right] = \frac{{2Qqa}}{{4\pi {\varepsilon _0}\left( {{a^2} - {x^2}} \right)}} \cr & {U_i} - {U_f} = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{a} - \frac{a}{{\left( {{a^2} - {x^2}} \right)}}} \right] \cr & = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{a^2} - {x^2} - {a^2}}}{{a\left( {{a^2} - {x^2}} \right)}}} \right] = \frac{{ - 2Qq{x^2}}}{{4\pi {\varepsilon _0}{a^3}}} \cr} $$
when $$x < < a$$ then $${x^2}$$ is neglected in denominator.
$${U_i} - {U_f} = \left( {\frac{{ - Qq}}{{2\pi {\varepsilon _0}{a^3}}}} \right){x^2}$$
