Question

5. Two positive charges A and B kept in air 0.1m apart experience a force of 16.2N. If the
charge B is 6uC, find the charge A. If another charge C of 2uC is placed between A and
B at a distance of 0.04m from A, find the resultant force on the charge C.​

Answer 1

Two positive charges A and B kept in air 0.1 m apart experience a force of 16.2N. the

charge B is 6uC. also another charge C of 2μC is placed between A and B at a distance of 0.04 m from A.

To find : (1) The charge A.

(2) the resultant force on the charge C.

solution : electrostatic force, F = kqQ/r²

here q = 6μC = 6 × 10¯⁶ C , r = 0.1 m and F = 16.2 N

so, 16.2 = (9 × 10^9 × 6 × 10¯⁶ × Q)/(0.1)²

⇒16.2 × 10¯² = 54 × 10³ × Q

⇒Q = (16.2/54) × 10¯⁵ = 3 × 10¯⁶C = 3μC

Therefore charge B is 3μC

now another charge C is placed between A and B at a distance of 0.04 m from A.

force between A and C, F₁ = kqq'/r'²

= (9 × 10^9 × 6 × 10¯⁶ × 2 × 10¯⁶)/(0.04)²

= 108 × 10¯³/16 × 10¯⁴

= 108/16 × 10

= 67.5 N (along AB )

force between C and B , F₂ = kq'Q/(0.1 - r')²

= (9 × 10^9 × 2 × 10¯⁶ × 3 × 10¯⁶)/(0.06)²

= (54)/(36) × 10¹

= 540/36

= 15 N (along BA)

resultant force on the charge C = F₁ - F₂ (both forces are in opposite directions so sign must be negative)

= 67.5 N - 15 N

= 52.5 N

Therefore the resultant force on the charge C is 52.5 N.