Hi! I\'m really not sure how to answer this question, can anyone please help? Th

Question

Hi! I'm really not sure how to answer this question, can anyone please help? Thanks a ton!

While a capacitor is connected to a battery, the plate separation is halved, from d to d/2. State how this affects the following. Express answers in terms of the original value. The capacitor starts with no dielectric material between the plates. The plates are brought back to their original separation d with no dielectric between them, and then the battery is disconnected. Then, a dielectric of constant k = 3 is inserted, filling the space between the plates. State how this affects the following. Express answers in terms of the original value.

Explanation / Answer

The sign of unknown charge will be positive as potential energy is given positive and known charge is positive and P.E. = Kq1q2/r.

And you can also say that potential given at origin also greater than the potential due to known charge hence unknown charge has to be positive.

Unknown charge must be placed on +ve x axis in order to get electric field in -ve x direction at origin as electric field lines are away from the positive charge.

Assume unknown charge is Q' and located at +x from origin.

P.E. = KQQ'/(x+d) = 2kQ2/d

Potential at origin KQ/d + KQ'/x = 4KQ/d

Solving both equations you will get x = 2d and Q' = 6Q.

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