When positive test charge is released from rest near (fixed) positive source cha

Question

When positive test charge is released from rest near (fixed) positive source charge, what happens to the electric potential of the positive test charge? It will increase because the charge will move in the direction of the electric field. It will decrease because the charge will move in the direction opposite to the electric field. It will decrease because the charge will move in the direction of the electric field. It will remain constant because the electric field is uniform. It will remain constant because the charge remains at rest. When negative test charge is released from rest near (fixed) positive source charge, what happens to the electric potential of the negative test charge? It will increase because the charge will move in the direction of the electric field. It will increase because the charge will move in the direction opposite to the electric field. It will decrease because the charge will move in the direction of the electric field. It will remain constant because the electric field is uniform. It will remain constant because the charge remains at rest. If we look at the electric potential around negative point charge, the potential will be Greater as we get closer to the charge Greater as we get farther from the charge Equal every where Its more complicated than that have no idea

Explanation / Answer

(1) when a positive test charge is placed near a positive soure charge ,the test charge will be replled and will hence move in the direction of electric field and away from positive source charge. so the electric potential decreases.so option (c) is correct.

(2)when a negative test charge is placed near a positive soure charge ,the test charge will be attarcted and will hence move in the direction opposite to that of electric field and towards the positive source charge. so the electric potential icreasess.so option (b) is correct.

(3) eletrci potential due to a negative charge is v= -kq/r

as r is increased, v will increase . (magintude will decrease but v is negative, so V i actuall increasing).so option (b )is correct

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