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## 400. Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10-8C) a) 4 b) 2 c) 8 d) 6

a Explanation: E = Q/ (4∏εor2)Q = (4000 X 0.32)/ (9 X 109) = 4 X 10-8 C.

a

See lessExplanation: E = Q/ (4∏εor2)Q = (4000 X 0.32)/ (9 X 109) = 4 X 10-8 C.

## 399. What is the electric field intensity at a distance of 20cm from a charge 2 X 10-6 C in vacuum? a) 250,000 b) 350,000 c) 450,000 d) 550,000

c Explanation: E = Q/ (4∏εor2)= (2 X 10-6)/(4∏ X εo X 0.22) = 450,000 V/m

c

See lessExplanation: E = Q/ (4∏εor2)= (2 X 10-6)/(4∏ X εo X 0.22) = 450,000 V/m

## 398. Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air. a) 18 X 109 b) 9 X 109 c) 36 X 109 d) -18 X 109

b Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109E = F/q = 18 X 109 /2 = 9 X 109 .

b

See lessExplanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109E = F/q = 18 X 109

/2 = 9 X 109

.

## 397. Find the force on a charge 2C in a field 1V/m. a) 0 b) 1 c) 2 d) 3

c Explanation: Force is the product of charge and electric field. F = q X E = 2 X 1 = 2 N.

c

See lessExplanation: Force is the product of charge and electric field.

F = q X E = 2 X 1 = 2 N.

## 396. The electric field intensity is defined as a) Force per unit charge b) Force on a test charge c) Force per unit charge on a test charge d) Product of force and charge

c Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2)

c

See lessExplanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2)

## 395. For a charge Q1, the effect of charge Q2 on Q1 will be, a) F1 = F2 b) F1 = -F2 c) F1 = F2 = 0 d) F1 and F2 are not equal

b Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.

b

See lessExplanation: The force of two charges with respect with each other is given by F1 and

F2. Thus F1 + F2 = 0 and F1 = -F2.

## 394. A charge of 2 X 10-7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10-7 C, both the charges are in vacuum. a) 0.03 b) 0.05 c) 0.07 d) 0.09

d Explanation: F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) = We get r = 0.09m.

d

See lessExplanation: F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) =

We get r = 0.09m.

## 393. Two small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving. a) 0.5 b) 0.4 c) 0.3 d) 0.2

c Explanation: F = mg = 10 X 10-3 X 9.81 = 9.81 X 10-2 N. On calculating r by substituting charges, we get r = 0.3m.

c

See lessExplanation: F = mg = 10 X 10-3 X 9.81 = 9.81 X 10-2 N.

On calculating r by substituting charges, we get r = 0.3m.

## 392. The Coulomb law is an implication of which law? a) Ampere law b) Gauss law c) Biot Savart law d) Lenz law

: b Explanation: The Coulomb law can be formulated from the Gauss law, using the divergence theorem. Thus it is an implication of Gauss law.

: b

See lessExplanation: The Coulomb law can be formulated from the Gauss law, using the

divergence theorem. Thus it is an implication of Gauss law.

## 391. Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN. a) 1.44 b) 2.44 c) 1.404 d) 2.404

c Explanation: Before the charges are brought into contact, F = 11.234 μN. After charges are brought into contact and then separated, charge on each sphere is, (q1 + q2)/2 = 0.5nC On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN

c

See lessExplanation: Before the charges are brought into contact, F = 11.234 μN.

After charges are brought into contact and then separated, charge on each sphere is,

(q1 + q2)/2 = 0.5nC

On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN