Question

OL: When two identical charges placed 1m apart the electric force between them is 9 x 10 N. The magnitu of each charge (in C) is: a) 10 92: The reading of the ammeter, shown in the circuit, -6 b) 10 c) 10 d) 10 e) 10 will be 5.2 A if point A is connected with: a) points B, C, and D b) point B and C only c) point B only d) point B and D only e) point C and D only 3Ω 2 6 2 C A 2Ω 12Ω

Answer 1

(1) Suppose magnitude of each charge = q Coulomb

Now, expression for the force between the two charges is –

F = k*q1*q2 / r^2------------------------------------(i)

where, k = Coulomb’s constant = 9.0 x 10^9 N*m^2/C^2

q1 = q2 = q

r = 1.0 m (given)

F = 9.0 x 10^15 N

Put these values in equation (i) –

9.0 x 10^15 = (9.0 x 10^9 x q x q) / 1.0^2

=> 9.0 x 10^15 = 9.0 x 10^9 x q^2

=> q^2 = (9.0 x 10^15) / (9.0 x 10^9) = 1.0 x 10^6

=> q = 1.0 x 10^3 C

So, option (a) is the correct answer.

(2) Required current, I = 5.2 A

So, resistance across the circuit should be, R = V / I = 28 / 5.2 = 5.38 Ohm

Resistance 1 ohm and 2 ohm are in series.

So, the required effective resistance = 5.38 – (1+2) = 2.38 ohm

Now, when the resistance 3 ohm and 12 ohm are connected in parallel.

Effective resistance, R = (3*12) / (3+12) = 2.40 ohm

This value is approximately equal to 2.38 ohm.

Therefore, point A should be connected to point B and D.

Hence, option (d) is the correct answer.