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  • . Two equal point charges are fixed at  <img alt="x=-a" src="https://learn.careers360.com

. 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

Option: 1

X


Option: 2

x^2


Option: 3

x^3


Option: 4

1/x


Answers (1)

best_answer

As we have learned

Potential of a System of Charge -

V=\frac{kq_{1}}{r_{1}}+k\frac{q_{2}}{r_{2}}+\frac{k\left ( -q_{3} \right )}{r_{3}}+\cdots

- wherein

 

 

 

 

 

Initially according to figure (i) potential energy of Q is U_i= \frac{2kqQ}{a}           ......(i)

 

According to figure (ii) when charge Q is displaced by small distance x then it’s potential energy now

U_f= kqQ\left [ \frac{1}{a+x}+\frac{1}{a-x} \right ]=\frac{2kqQa}{a^2-x^2}....(2)

 

Hence change in potential energy

\Delta U= U_f-U_i= 2kqQ\left [ \frac{a}{a^2-x^2}+\frac{1}{a} \right ]=\frac{2kqQx^2}{a^2-x^2} 

Since x<<a so \Delta U=\frac{2kqQx^2}{a^2}\Rightarrow \Delta U\propto x^2

 

 

Posted by

vishal kumar

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