A positive point charge q = +2.58 nC is located at x = 1.20 m and a negative cha

Question

A positive point charge q = +2.58 nC is located at x = 1.20 m and a negative charge of

?2q = ?5.16 nC is located at the origin as in the following figure.

(a) Sketch the electric potential versus x for points along the x-axis in the range ?1.50 m < x < 1.50 m.

This answer has not been graded yet.

(b) Find a symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges. (Use the following as necessary: x.)
V = ____________________________________________

(c) Find the electric potential at x = 0.576 m.


(d) Find the point along the x-axis between the two charges where the electric potential is zero.

Please show work I am attempting to learn this thank you!!

Explanation / Answer

(b)

We can see from the graph that potential due to positive charge is +ve and that due to negative charge is -ve.

So, the potential at any point between the points where the two charges are placed is given by the sum of the effects of the two charges -2q and q.

Let me take 1/(4*pi*epsilon) = K

Then,

Potential at point P due to charge -2q, P1 = K *(-2q)/x

& Potential at point P due to charge q, P2= K*q/ (1.20-x)

where x is the distance from the charge -2q.

So the total potential is the superposition of the two potentials

i.e. P(final)= P1 + P2 = K *(-2q)/x +   K*q/ (1.20-x) ............. (1)

(c)

The electric potential at point P when x= 0.576m can be obtained by putting x= 0.576 in the expression (1)

The answer will be negative because the point is closer to the higher magnitude -ve charge.

(d)

It should be very clear that the point between -2q and q where effective potential is zero is going to be closer to the charge of lower magnitute. Because then that charge will get enough lower x to make the value of the potential cancel the potential of the higher charge.

Putting the above expression equal to zero.

x= 0.8 will be obtained, after doing a little algebra.

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