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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