abhishek singh
Last Activity: 10 Years ago
answer is B, D
see as we know that , charge on the facing surfaces of plates of a capacitors are same in magnitude but opposite in polarity , and those of outer surfaces are same in magnitude as well as polarity .
using this concept , ,,
before the connection of battery let the charge on plate A be 100pC and that on B be 200pC , , now let the inner surface of plate A have a charge x pC and the outer surface of plate A will have (100-x)pC charge . therfore the inner surface of plate B will have a charge -x pc . and outer surface of this plate will have a charge ( 200+x )pC.
now using the above concept ,,,,, 100-x = 200 + x => x = -50pC ------ C option is correct
After the battery is connected … let the inner surface of plate A have a charge y pC and outer surface have a charge Q . therfore the inner surface of plate B will have a charge -y pC and outer surface have a charge Q . Also these plates have a potential difference of 10v ,, therfore y/c = v => y = 50pC .. ------- option D is wrong
NOW WE NEED TO APPLY CHARGE CONSERVATION . i.e total initial charge = total final charge
therfore 100+200 = Q + Q + y +( -y) => Q = 150
therfore battery has supplied a charge of 100 p C ---- option A is correct . now enerf supplied = charge supplied by battery x e.m.f . i.e 10x 100x10-6 j = 10-2 j---option B is wrong
PLEASE TELL ME IF MY ANSWER OR SOLUTION IS WRONG
