Question

Two point charges q1=3×10−6C and q2=5×10−6C are located at (3,5,1) and (1,3,2)m . Find the force on q1 due to q2.

Answer 1

Given:

q1=3×10−6C is located at (3,5,1) and

q2=5×10−6C is located at (1,3,2).

To Find:

The force on q1 due to q2.

Solution:

Let R be the distance between q1 and q2.

Then force on q1 due to q2 is given by,

• F = $\frac{1}{4\pi \epsilon_0}$ q1 q2 /R²
• q1 q2 = 3x $10^{-6}$ x 5 x $10^{-6}$ = 15x$10^{-12}$
• R = distance between q1 and q2.
• R = $\sqrt{(3-1)^{2} +(5-3)^{2}+ (1-2)^{2} }$ = $\sqrt{2^{2} +2^{2}+1^{2} }$ = 3

Therefore force on q1 due to q2 ,

• F = 9 x $10^{9}$ x 15 x $10^{-12}$ / 3²

• F = 15 x $10^{-3}$ N

The force on q1 due to q2 is F = 15 x $10^{-3}$ N.