electric field strength at a point between oppositely charged plates is E if the distance between the plates is reduced to half what will be the value of new electric intensity?
Can it be 4E as potential will also double as distance halves?
Answers
Answered by
1
Explanation:
the answers... i think all are correct
Attachments:
Answered by
5
Explanation:
1.Electric field remains unchanged as we reduce the distance to half because electric field is independent of distance between the plates.
electric field intensity is given as:
E = Q/AE°
where E= electric field
Q= charge between plates
A= Area of the plates
E°= epsilon
so, new electric field is E.
2. Yes, if we double the potential and distance is halved then Electric field will become 4E.
U = Q^2/2C
where, U= potential energy
Q= Charge
C= capacitance
refer attachment......
follow me for getting more conceptual answer...... nd mark as brainliest...
THANK U
Attachments:
Similar questions