A charge of #-1 C# is at the origin. How much energy would be applied to or released from a # -3 C# charge if it is moved from # ( -1, -3 ) # to #(-2 ,7 ) #?

1 Answer
Jan 23, 2016

Answer:

#0.179kcolor(white)(x)"units"#

Explanation:

MFDocs

The -3C charge moves from #P# to #Q#.

I will work out the electric potential due to the charge at the origin at points #P# and #Q#.

Then I will subtract these to get the potential difference between the 2 points in volts

If 1C of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

So I will then multiply the potential difference by -3C to get the total work done.

Potential at P:

This is given by:

#E_P=(k.q)/d_1#

#k# is a constant.

To find #d_1# we use Pyhtagoras:

#d_1=sqrt(2^2+7^2)=sqrt(53)#

#d_1=7.28#

#:.E_P=(kxx-1)/7.28=-k/7.28" "color(red)((1))#

Potential at Q:

#E_Q=k.q/d_2#

Again, we can use Pythagoras to find #d_2rArr#

#d_2=sqrt(1^2+3^2)=sqrt(10)#

#d_2=3.16#

#:.E_Q=(kxx-1)/3.16=(-k)/3.16" "color(red)((2))#

To get the potential difference #DeltaE# between the 2 points we subtract #color(red)((1))# from #color(red)((2))rArr#

#DeltaE=-k/7.28-(-k/3.16)#

#DeltaE=k[1/3.16-1/7.28]#

#DeltaE=k[0.316-0.137]=0.178k#

To get the work done #W# we multiply by the charge moved:

#W=-3xx0.178k=-0.536kcolor(white)(x)"units"#

The negative sign means work must be done on the charge to move it from #P# to #Q#.

No units for distance are given. If it was in metres then #k=9xx10^9"m/F"# and the answer would be in Joules.