Two large, parallel conducting plates X and Y, kept close to each other, are given charges Q1 and Q2(Q1>Q2). The four surfaces of the plates are A, B, C and D, as shown in figure. Then :
This question has multiple correct options
The charge on A is 21(Q1+Q2).
The charge on B is 21(Q1−Q2).
The charge on C is -21(Q1−Q2).
The charge on D is 21(Q1+Q2)
Open in App
Updated on : 2022-09-05
Verified by Toppr
Correct options are A) , B) , C) and D)
Let charge for surface A =Q1−q, for surface B =q, for surface C =Q2−q′ and for surface D =q′. As the field inside X plate is zero, 2Aϵ01(Q1−q−q−Q2+q′−q′)=0 −2q=Q2−Q1⇒q=21(Q1−Q2) similarly for Y plate, 2Aϵ01(q′−Q2+q′−q−Q1+q)=0 or 2q′=Q1+Q2⇒q′=21(Q1+Q2) Thus, charge on A is Q1−q=Q1−21(Q1−Q2)=21(Q1+Q2). charge on B is q=21(Q1−Q2) charge on C is Q2−q′=Q2−21(Q1+Q2)=−21(Q1−Q2) Charge on D is q′=21(Q1+Q2)