18 - Capacitors Questions Answers
distance b/w plate of capacitor C &q charge is increased to double . work done=??
as distance d is double thus C= Ae/2d thus capacitance will be halved so wd shld be (1/2 )q2/(C/2) ie 1/4 q2/C
BUT Sir the ans given is=( 1/2)q2/C Plzz do me a little favour by solving this query
(1/2 )q2/(C/2) = q2/C not 1/4 q2/C
so work done = new energy - previous energy
a II plate capacitor has an electric field of 105 V/m b/w d plates . if charge on capacitor plate is 1uC d force on each capacitor is??
ans 0.05N
F = σq/2ε0 = Eq/2 = 105*10-6/2 = 0.05
A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates. Please add explanation!
due to induction the charges on the four faces will be
so V = q/C = 1.5 * 10-8/1.2 * 10-9
solve now
according to the diagram
tanθ = qE/mg = q(σ/2ε0)/mg
solve this eq. for σ.
Remember that length of the thread is not applicable in the question.