Question 843: Electrostatics & Capacitance

Question

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

Accepted Answer

Number of electric field lines originating from Q
1
is more than terminating at Q
2
.
$$	herefore$$ $$|{Q_1}| > |{Q_2}|$$
Here, Q
1
is positive while Q
2
is negative.
The electric field at a distance x towards the right of Q
2
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 Q
1
and Q
2
. Since $${Q_1} > {Q_2}$$, $$|\overrightarrow E |$$ becomes zero for some finite x.

Answer

$$\left| {{Q_1}} \right| > \left| {{Q_2}} \right|$$

Answer

$$\left| {{Q_1}} \right| < \left| {{Q_2}} \right|$$

Answer

at a finite distance to the left of $${{Q_1}}$$ the electric field is zero

Answer

at a finite distance to the right of $${{Q_2}}$$ the electric field is zero

Electrostatics & Capacitance

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

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