# 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?

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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#