A 2 V battery is connected to two parallel plates a distance 55 cm apart. If a #-4.30 xx 10^-4 C# charge was placed in the electric field, what force would it feel?
What work is needed to move the #-4.30 xx 10^-4 C# charge from 0.138 m from the ground plate to 0.413 m from the ground plate?
What work is needed to move the
1 Answer
Electric field is given by
#vecE=V/dhat r#
direction of#hat r# is from positive plate to negative plate.
Inserting given values we get
Force experienced by a charged particle
#vecF=qvecE#
Hence , using (1) we get
#vecF=2/0.55xx(-4.30xx1^-4)hat r#
#|vecF|=1.56xx10^-3N#
Direction is opposite to that of#hat r# , i.e., towards positive plate
Work done in moving the charge is
#W=vecFcdot vecs#
As the angle between displacement and force vectors is
#:.W=|vecF| |vecs|#
#=>W=1.56xx10^-3xx (0.413-0.138)#
#=>W=1.56xx10^-3xx (0.413-0.138)#
#=>W=4.3xx10^-4J#