An IDEAL parallel plate capacitor with plate separation d is

An IDEAL parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V Volts. (C is the capacitance, U is the stored energy, and E is the magnitude of the electric field inside the capacitor.) (For each statement select T True, F False). A) With the capacitor connected to the battery, decreasing d has no effect on the E. B) After being disconnected from the battery, decreasing d increases U. C) With the capacitor connected to the battery, increasing d increases C. D) After being disconnected from the battery, increasing d increases V.

Solution

A. TRUE

If we try getting the resultant field using Gauss\'s Law, enclosing the plate in a Gaussian surface as shown, there is flux only through the face parallel to the positive plate and outside it (since the other face is in the conductor and the electric field skims all other faces).sice charge inclose inside guassian surface is same for both the plate so electric filed remains same or no effect on E.

B.)

TRUE,As d is decreases U will icrease

U = 1/2*C*V^2
where:
C= capacitance (in Farads)
V = Voltage
Q = Coulombs
U = energy (in joules)

C= k*epsilon0*A/d

U = 0.5 *k*A(C= k*epsilon0*A^d 2/epsion*d

C.)False

, increasing d decreases C. Being connected to battery does not matter.

D.)TRUE

after being disconnected from battery charge in capacitor is constant and C decreases when d increases so V will increases

Q = CV

C = kEpsilon A/d