ABC is a right triangle in which AB = 3 cm, BC = 4 cm and right angle is at B. Three charges `+ 15mu C + 12 mu C` and `-20 mu C` are placed respectively at A, B and C. The force acting on the charge at B is

Options

A. 1250 N
B. 3500 N
C. 1200 N
D. 2250 N

Answer

D

Explanation

According to the question, 3 charge particles are placed at the vertices of a right angle triangle ABC as shown in the figure below,

Where, `Q_(A) = +15 mu C`,
`Q_(B) = 12 mu C`
and `Q_(C ) = -20 mu C`
Now, the force, `F_(AB) = (kQ_(A)Q_(B))/(r_(AB)^(2))`
`rArr F_(AB) = (k15 xx 12 xx 10^(-12))/(9 xx 10^(-4)) = k 20 xx 10^(-8) N`
Similarly,
`F_(BC) = (k xx 20 xx 12 xx 10^(-12))/(16 xx 10^(-4)) = k15 xx 10^(-8) N`
Now, the resultant, `F_(B) = sqrt(F_(AB)^(2) + F_(BC)^(2)) " "(because theta = 90^(@))`
`F_(B) = 9 xx 10 [sqrt(20^(2) + 15^(2)] = 2250 N`
Hence, the force acting on the charge at point B is 2250 N.
So, the correct option is (d).