wiz-icon
MyQuestionIcon
MyQuestionIcon
11
You visited us 11 times! Enjoying our articles? Unlock Full Access!
Question

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

A
V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
V+QC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
V+Q2C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
VQC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C V+Q2C
Let the area of the plate be A and distance between the plates be d.

Initial charge on the plates,

Q0=CV

As, C=ε0Ad

Q0=Aε0dV .....(1)

Now, when charge Q is added to the positive plate, electric field between the plates will be,

E=Q+Q02Aε0+Q02Aε0



Now new voltage across the plates will be,

V=Ed=Qd+Q0d2Aε0+Q0d2Aε0

V=Qd2Aε0+Q0d2Aε0+Q0d2Aε0

V=Qd2Aε0+Q0dAε0

From equation (1), V=Q0dAε0 and C=ε0Ad, we have

V=Q2C+V

Hence, option (c) is the +correct answer.
Key concept : Effect on potential difference on PPC (Parallel Plate Capacitor) with battery removed after charging and when some more charge is given to its positive plate after removal of battery.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon