Question

Two point charges q1 and q2 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

• A. 16
• B. 8
• C. 1
• D. 2

Answer 1

Answer: (A)

16

Answer 2

Let's assume the original charges are q1 and q2, and the distance between them is l. The initial force is given by:
Finitial = k * $\frac {|q_1 * q_2|}{l^2}$
Now, if one of the charges is doubled (let's say q1) and the distance between them is halved ($\frac {1}{2}$), the new force is given by:
Fnew = k * $\frac {|2q_1 * q_2|}{(\frac {l}{2})^2 }$
Fnew = 4 * k * $\frac {|q_1 * q_2|}{(\frac {l}{2})^2 }$
Fnew = 4 * k * $\frac {|2q_1 * q_2|}{(\frac {l^2}{4}) }$
Fnew = 16 * k * $\frac {|2q_1 * q_2|}{l^2}$
Comparing the new force to the initial force:
$\frac {F_{new}}{F_{initial} }$ = $\frac {16*k*|q_1*q_2| / l^2|}{k*|q_1*q_2| / l^2}$
$\frac {F_{new}}{F_{initial} }$ = 16
Therefore, the magnitude of the force becomes 16 times the initial force.
The answer is (A) 16.