# Derive an expression for V(x) / (keQ / a). (Use the following as necessary: x and a.)

Two point charges of equal magnitude are located along the y axis equal distances above and below the x axis, as shown in the figure below. Note: Assume a reference level of potential V = 0 at r = .

(a) Derive an expression for V(x) / (keQ / a). (Use the following as necessary: x and a.)

 V(x) (keQ/a)

=

Plot a graph of the potential at points along the x axis over the interval -3a < x < 3a. You should plot the potential in units of keQ/a. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Derive an expression for V(y) / (keQ / a). (Use the following as necessary: y and a.)

 V(y) (keQ/a)

=

Let the charge located at -a be negative and plot the potential along the y axis over the interval -4a < y < 4a. (Do this on paper.

asked Jan 20, 2012 in Physics

First start by noting that V(x) = (ke*Q1) / (r1) + (ke*Q2) / (r2)  now subtstiute in values to get

that  V(x) = ( ke(+Q) / (x^2 + a^2)^(1/2)) + ( ke(+Q) / (x^2 + (-a)^2)^(1/2))

and simplifying we get that:

V(x) = ( 2ke(Q) / (x^2 + a^2)^(1/2)) =   ke*Q / a * (2 /  ((x/a)^2 + 1))^(1/2)

Now divide V(x) by keQ/a to get that

V(x) / (keQ / a) =  2 / ((x/a)^2 + 1 )^(1/2)

Similarly for part B, establish that V(y) = (ke*Q1) / (r1) + (ke*Q2) / (r2) which is equal to:

V(y) = ( ke(+Q) /  | y - a | ) +  ( ke(-Q) /  | y + a | )

now simplify a bit to get:

V(y) = ke*Q / a ( 1/ | y/a -1 |  -  1/ | y/a +1 |)

and so

V(y) / (keQ / a) =  ( (1/  | y/a -1 |)  -  (1/  | y/a +1 |))

(*note that | | are absolute value signs)

answered Jan 26, 2012 by ~Expert~ (3,020 points)
selected Jan 26, 2012 by Joey33