A charge of #5 C# is at the origin. How much energy would be applied to or released from a # -4 C# charge if it is moved from # ( 8, 2 ) # to #(9 , 6 ) #?

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Answer 1 · 2015-12-30T00:03:13

#0.43k# units of work applied to the charge.

Click here for the situation:

We can work out the electric field potential at point R.

Then we work out the electric field potential at point Q.

We can then get the work done in moving a charge of #-4C# between these 2 points by multiplying the charge by the potential difference.

Potential at point R:

#V_R=-(k.5)/r" "color(red)((1))#

From the diagram you can see that:

#r^2=8^2+2^2=68#

#:.r=sqrt(68)=8.24#

So from #color(red)((1))#:

#V_R=-k.5/8.24=-k.0.607#

Potential at point Q:

#V_Q=-(k.5)/d" "color(red)((2))#

From the diagram you can see that:

#d^2=8^2+6^2=100#

#:.d=sqrt(100)=10#

So from #color(red)((2))#:

#V_Q=-(k.5)/10=-k/2#

So the potential difference #E# between R and Q #rArr#

#E=-k0.607-(-k0.5)#

#E=-0.107k#

The work done in bringing a charge of #-4C# through this potential difference is given by:

#W=-0.107kxx(-4)#

#:.W=0.43k# units

This means work would have to be applied to the charge.

There are no units of distance given. If you were using metres then #k=9xx10^(9)"m/F"# and the answer would be in Joules.