Question

A positively charged ball hangs from a silk thread. Electric field at a certain point (at the same horizontal level of the ball) due to this charge is $E$. Let us put a positive test charge ${{q}_{o}}$ at a point and measure $\dfrac{F}{{{q}_{o}}}$, then it can be predicted that the electric field strength $E$.

A) $>\dfrac{F}{{q}_{o}}$

B) $=\dfrac{F}{{q}_{o}}$

C) $< \dfrac{F}{{q}_{o}}$

D) Cannot be estimated

Answer 1

Hint: The positively charged ball hanging on the thread will produce an electric field and the charge on the test charge will experience a force equal to the product of the charge on it and the magnitude of the electric field it is in. Using this information, we can find the electric field strength.

Formula used:

Coulomb’s Law: $F=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Complete step-by-step solution:

In the question, we are given a positive charged ($q$) ball hanging by a silk thread let the electric field due to this charge is given by,

$E=\dfrac{kq}{r^2}$

Now we have put positive test charge ${{q}_{o}}$ at the same horizontal level of the ball and ball is also have positive charge so there will be a repulsive force between these charges due to which the distance between them get increased by $r’$ and the force between them is given by,

$F=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{q{{q}_{o}}}{{{r’}^{2}}} $

$\Rightarrow \dfrac{F}{q_o}=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{q}{{{r’}^{2}}}$

Since $r’ > r$ so the value of $\dfrac{F}{q_o}$ will decreased and hence we can write,

$\therefore E>\dfrac{F}{q_o}$

Hence, the correct option will be (A).

Note: Take care that here we have been asked about the electric field strength and not the electrostatic force. The ratio of the electrostatic force and the charge will give the strength of the electric field that the charge is present in.