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A positive charge q is fixed at the point x=0,y=0 and a negative charge -2q is fixed at the point x=a,y=0.

Part A:

Derive an expression for the potential V at points on the y-axis as a function of the coordinate y. Take V to be zero at an infinite distance from the charges.

Part B:

At which positions on the y-axis is V = 0?

Part C:

What does the answer to part A become when y>>a?

Part D:

Explain why this result is obtained.

Added Thu, 02 Jul '15

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Solutions

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A positive charge 'q' is fixed at the point x= 0, y= 0, and a negative charge '-2q' is fixed at the point x=a , y = 0.

a) The electric potential V is a scalar

Suppose P is a point at distance y from +q on the x axis

The electric potential at distance x from+q =V1= kq / y

The electric potential at distance (y-a) from - 2q =V2= -2kq / (y - a)

Total electric potential =V1 + V2

Total electric potential = kq / y -2kq / (y - a)

Total electric potential =kq { (1 / y) - [2 / (y - a)] }

Total electric potential =kq ( y -a -2y) / y (y - a

Total electric potential =kq ( y + a) / x (a - y)

= (1/4pi eo)[1/y -2/(y-a)

_____________________________________

b) V = 0 at positions y/a = -1,0.333 , on the -axis

_______________________________

c)when y >>a the answer to part A becomes V = -ky /a

______________________________________...

At large 'y' , the potential due to -2q is nearly twice the potential due to +q , hence resultant potential is due to sum of - 2q and +q which is - q

Potential at 'y' = - ky/a

Added Thu, 02 Jul '15

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