JEE Advance - Physics (2010 - Paper 1 Offline - No. 9)
A few electric field lines for a system of two charges $${Q_1}$$ and $${Q_2}$$ fixed at two different points on the $$x$$-axis are shown in the figure. These lines suggest that
$$\left| {{Q_1}} \right| > \left| {{Q_2}} \right|$$
$$\left| {{Q_1}} \right| < \left| {{Q_2}} \right|$$
at a finite distance to the left of $${{Q_1}}$$ the electric field is zero
at a finite distance to the right of $${{Q_2}}$$ the electric field is zero
Explanation
Number of electric field lines originating from Q1 is more than terminating at Q2.
$$\therefore$$ $$|{Q_1}| > |{Q_2}|$$
Here, Q1 is positive while Q2 is negative.
The electric field at a distance x towards the right of Q2 is given by
$$|\overrightarrow E | = {1 \over {4\pi { \in _0}}}{{{Q_1}} \over {{{(d + x)}^2}}} - {1 \over {4\pi { \in _0}}}{{{Q_2}} \over {{x^2}}}$$,
where d is the separation between Q1 and Q2. Since $${Q_1} > {Q_2}$$, $$|\overrightarrow E |$$ becomes zero for some finite x.
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