Q20. Electric Charge Force and Field

A small object has charge Q A small object has charge Q. Charge q is removed from it and placed on a second small object. The two objects are placed 1 m apart. For the force that each object exerts on the other to be a maximum, q should be : 2 Q Q Q / 2 Q / 4 <PowerClick><Answer>3</Answer><Option>5</Option></PowerClick>

Q  q q  ( max )

Two identical conducting spheres A and B carry equal charge Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes : F / 2 F / 4 3F / 8 F / 16 <PowerClick><Answer>3</Answer><Option>5</Option></PowerClick>

Spheres identical  Charges are shared equally after contact A, Q B, Q C, 0 Spheres identical  Charges are shared equally after contact After C touching A : After C touching B : Final force between A & B :

Particles 1, with charge q1, and 2, with a charge q2, are on the x axis, with particle 1 at x = a and particle 2 at x  –2a. For the net force on a third charged particle at the origin to be zero, q1 and q2 must be related by q2  : 2 q1 4 q1 – 2 q1 – 4 q1 – q1 / 4 <PowerClick><Answer>2</Answer><Option>5</Option></PowerClick>

2a q a q2 q1 

A particle with charge 2 C is placed at the origin A particle with charge 2 C is placed at the origin. An identical particle, with the same charge, is placed 2 m from the origin on the x axis, and a third identical particle, with the same charge, is placed 2 m from the origin on the y axis. The magnitude of the force on the particle at the origin is 9.0  10–3 N 6.4  10–3 N 1.3  10–2 N 1.8  10–2 N 3.6  10–2 N <PowerClick><Answer>3</Answer><Option>5</Option></PowerClick>

y = 2 m q = 2C x = 0 q = 2C x = 2 m q = 2C

just use R for the distance just use 2R for the distance A charge Q is spread uniformly along the circumference of a circle of radius R. A point particle with charge q is placed at the center of this circle. The total force exerted on the particle q can be calculated by Coulomb's law : just use R for the distance just use 2R for the distance just use 2p R for the distance result of the calculation is zero none of the above <PowerClick><Answer>4</Answer><Option>5</Option></PowerClick>

Q q R f  0 by symmetry.

A particle with charge Q is on the y axis a distance a from the origin and a particle with charge q is on the x axis a distance d from the origin. The value of d for which the x component of the force on the second particle is the greatest is : a a  2 a / 2 a /  2 <PowerClick><Answer>5</Answer><Option>5</Option></PowerClick>

Q r a f d q 

A particle with a charge of 5  10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a stationary particle with a charge of 5  10–6 C. The radius of the orbit is : 0.23 m 0.62 m 1.6 m Orbit is impossible <PowerClick><Answer>5</Answer><Option>5</Option></PowerClick>

Circular motion requires a centripetal force. Q = 5  10–6 C m = 20 g v = 7 m/s q = 5  10–6 C R Circular motion requires a centripetal force. Coulomb force between the particles is however repulsive, and hence centrifugal.