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anastassius [24]
3 years ago
12

A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnit

ude, the balls will be attracted to each other.
a. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.

b. The negative charges on the sphere A move to the side closer to sphere B, while the positive charges move to the side further away from sphere B. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.

c. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

d. The negative charges on the sphere B move to the side closer to sphere A, while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Physics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

D

Explanation:

Ball A is a non positively charged non metal while ball B is metal ball.

Given: The ball B positive charge of small magnitude

To prove: Balls will attract each other

IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

Hence the correct answer will be D .

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(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
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(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
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(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
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If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
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E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

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