A charge of #7 C# is at the origin. How much energy would be applied to or released from a # 5 C# charge if it is moved from # (-3 , 7 ) # to #(-7 ,-2 ) #?

1 Answer
Jan 7, 2016

Answer:

#0.565"k"# units

Explanation:

MFDocs

I'll work out the electric potential at points P and Q.

Then I will use this to work out the potential difference between the 2 points.

This is the work done by moving a unit charge between the 2 points.

The work done in moving a 5C charge between P and Q can therefore be found by multiplying the potential difference by 5.

We need to find the length of PR and PQ which can be done using Pythagoras:

#PR^2=2^2+7^2#

#:.PR=sqrt(4+49)#

#PR=7.8#

and:

#QR^2=3^2+7^2#

#:.QR=sqrt(9+49)=6.93#

The electric potential due to a charge #q# at a distance #r# is given by:

#V=k.q/r#

#k# is a constant.

So the potential at point #P# is given by:

#V_P=kxx7/7.8=0.897k#

The potential at #Q# is given by:

#V_Q=kxx7/6.93=1.01k#

So the potential difference is given by:

#V_P-V_Q=k(0.897-1.01)=-0.113k#

So the work done in moving a 5C charge between these 2 points is given by:

#W=-0.113kxx5=-0.565k"units"#

This is the work done on the charge.

There are no units of distance given. If this was in meters then #k=9xx10^9"m/F"# and the answer would be in Joules.