Electrostatics 5 Question 30

32. Two identical metal plates are given positive charges Q1 and Q2(<Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is

(1999, 2M)

(a) (Q1+Q2)/2C

(b) (Q1+Q2)/C

(c) (Q1Q2)/C

(d) (Q1Q2)/2C

Show Answer

Solution:

  1. Electric field within the plates E=EQ1+EQ2

Potential difference between the plates

VAVB=Ed=Q1Q22Aε0d=Q1Q22Aε0d=Q1Q22C



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक