Question 767: Electrostats & Capacitance

Question

Two point charges $\mathrm{q}_1$ and $\mathrm{q}_2$ are ' $l$ ' distance apart. If one of the charges is doubled and distance between them is halved. The magnitude of force becomes $n$ times, where $n$ is

Accepted Answer

Two point charges $ q_1 $ and $ q_2 $ are $ l $ apart.
Step 1. Original force
By CoulombÃ¢ÂÂs law:
$ F = k \frac{q_1 q_2}{l^2} $
Step 2. Changes
One charge is doubled, say $q_1 	o 2q_1$.
Distance is halved: $l 	o 	frac{l}{2}$.
New force:
$ F' = k \frac{(2q_1)\,q_2}{\left(	frac{l}{2}ight)^2}. $
Step 3. Simplify
$ F' = k \frac{2 q_1 q_2}{	frac{l^2}{4}} = k \cdot (2 q_1 q_2) \cdot \frac{4}{l^2} = 8 \cdot \frac{k q_1 q_2}{l^2}. $
So
$ F' = 8F. $
Answer:
The new force is
8 times
the old force.
Correct option:
C (8)
.

Answer

1

Answer

2

Answer

8

Answer

16

Electrostats & Capacitance

Two point charges \mathrm{q}_1 and \mathrm{q}_2 are ' l ' distance apart. If one of the charges is doubled and distance between them is halved. The magnitude of force becomes n times, where n is

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