When a plastic rod is rubbed with silk, the plastic becomes positively charged.

Question

When a plastic rod is rubbed with silk, the plastic becomes positively charged. The silk becomes positive because it has gained protons. The silk becomes positive because it has lost electrons. The silk becomes negative because it has lost protons. The silk becomes negative because it has gained electrons. The silk becomes negative because it has lost electrons. Two equal charges q are separated by a distance L. The first charge is held fixed at rest, and the second charge is released. Find the final kinetic energy of the second charge when it has moved to a very large distance under the influence of the electric field produced by the first charge. KE_f = 0 KE_f = infinity KE_f = q^2/4 pi epsilon_0 L KE_f = q^2/8 pi epsilon_0 L q^2/4 pi epsilon_0 L^2 The figure shows four charges, q_1, q_2, q_3, and q_4. S is a Gaussian surface (closed surface) with q_1 and q_2 inside the surface. The charges q_3 and q_4 are not inside the surface. Consider Gauss' law and find the outward flux phi_E of the electric field through the surface S. phi_E = 0 phi_E = q_1 + q_2 - q_3 - q_4/epsilon_0 phi_E = q_1 + q_2 + q_3 + q_4/epsilon_0 phi_E = q_1 + q_2/epsilon_0 phi_E = q_3 + q_4/epsilon_0 A cube with sides of length L has a point charge Q exactly at one corner of the cube as shown. Find the outward electric flux through the surface of the cube.

Explanation / Answer

Part 1 A) when a plastic is rubbed with silk rod,the plastic acquire positive charge because it will lose some electrons,so silk will gain some electrons and acquire negative charge

so option D i.e silk becomes negative because it has gained electrons.

Part 1 B)

We define work done=change in kinetic energy

Work done= q(Vfinall-Vinitial)

Please note that Here V(capital V) is the potential

Vfinall=0 if we consider the other point at very large distance i.e at infinity

Now Vinitial=-k q/L where k=1/4o

therefore work done=(1/4o) q2/L

this work done is nothing but change in kinetic energy

work done=kE final- KEinitial

KEinitial= mvinitial2/2

vinitial=0 since the particle is moved from rest

therefore KEinitial= 0

which implies that

work done=kE final

therefore KEfinal= (1/4o) q2/L

So the answer is option C

Part 1 C) For a gaussian surface we will consider the charges which are present inside that surface ,and those charges will contribute towards the total flux

since q1 and q2 are present inside the gausian surface

Therfore total flux =(q1+ q2)/0

Option D is correct

GOOD LUCK!!!

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