A charge of #-2 C# is at the origin. How much energy would be applied to or released from a # 4 C# charge if it is moved from # (7 ,5 ) # to #(3 ,-2 ) #?

1 Answer
Jan 12, 2016

Let #q_1=-2C#, #q_2=4C#, #P=(7,5)#, #Q=(3.-2)#, and #O=(0.0)#
The distance formula for Cartesian coordinates is

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2#
Where #x_1, y_1#, and #x_2, y_2,# are the Cartesian coordinates of two points respectively.

Distance between origin and point P i.e #|OP|# is given by.

#|OP|=sqrt((7-0)^2+(5-0)^2)=sqrt(7^2+5^2)=sqrt(49+25)=sqrt74#

Distance between origin and point Q i.e #|OQ|# is given by.

#|OQ|=sqrt((3-0)^2+(-2-0)^2)=sqrt((3)^2+(-2)^2)=sqrt(9+4)=sqrt13#

Distance between point P and point Q i.e #|PQ|# is given by.

#|PQ|=sqrt((3-7)^2+(-2-5)^2)=sqrt((-4)^2+(-7)^2)=sqrt(16+49)=sqrt65#

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 two points.

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

The work done in moving a #4C# charge between #P# and #Q# can therefore be found by multiplying the potential difference by #4#.

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

#V=(k*q)/r#

Where #k# is a constant and its value is #9*10^9Nm^2/C^2#.

So the potential at point #P# due to charge #q_1# is given by:

#V_P=(k*q_1)/sqrt74#

The potential at #Q# due to the charge #q_1# is given by:

#V_Q=(k*q_1)/sqrt13#

So the potential difference is given by:

#V_Q-V_P=(k*q_1)/sqrt13-(k*q_1)/sqrt74=(k*q_1)(1/sqrt13-1/sqrt74)#

So the work done in moving a #q_2# charge between these 2 points is given by:

#W=q_2(V_Q-V_P)=4(k*q_1)(1/sqrt13-1/sqrt74)=4(9*10^9*(-2))(1/sqrt13-1/sqrt74)=-11.5993*10^9#

This is the work done on the charge.

There are no units of distance given. If this was in meters then the answer would be in Joules.