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

The Correct Option is A

Let's assume the original charges are q₁ and q₂, and the distance between them is l. The initial force is given by: 
F_initial = 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: 
F_new = k * \(\frac {|2q_1 * q_2|}{(\frac {l}{2})^2 }\)
F_new = 4 * k * \(\frac {|q_1 * q_2|}{(\frac {l}{2})^2 }\)
F_new = 4 * k * \(\frac {|2q_1 * q_2|}{(\frac {l^2}{4}) }\)
F_new = 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.

Answer 2

Initial Force (F_initial):
\(F_{\text{initial}} = k \frac{|q_1 q_2|}{l^2}\)

New Force (F_new) after changes:
One charge (q1) is doubled: \(2q_1\)
Distance between them is halved: \(\frac{l}{2}\)

\(F_{\text{new}} = k \frac{|2q_1 \cdot q_2|}{\left(\frac{l}{2}\right)^2}\)

\(F_{\text{new}} = k \frac{|2q_1 \cdot q_2|}{\frac{l^2}{4}}\)

\(F_{\text{new}} = 4 \cdot k \frac{|2q_1 \cdot q_2|}{l^2}\)

\(F_{\text{new}} = 16 \cdot k \frac{|q_1 \cdot q_2|}{l^2}\)

Comparing F_new to F_initial:
\(\frac{F_{\text{new}}}{F_{\text{initial}}} = \frac{16 \cdot k \frac{|q_1 \cdot q_2|}{l^2}}{k \frac{|q_1 \cdot q_2|}{l^2}}\)

\(\frac{F_{\text{new}}}{F_{\text{initial}}} = 16\)

So, the correct option is (A): 16