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

Question

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

b)At which positions on the x-axis is V = 0?

c)What does the answer to part A become when x>>a?

Explanation / Answer

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 x from +q on the x axis The electric potential at distance x from+q =V1= kq / x The electric potential at distance (x-a) from - 2q =V2= -2kq / (x - a) Total electric potential =V1 + V2 Total electric potential = kq / x -2kq / (x - a) Total electric potential =kq { (1 / x) - [2 / (x - a)] } Total electric potential =kq ( x -a -2x) / x (x - a) Total electric potential =kq ( x + a) / x (a - x) = (1/4pi eo)[1/x -2/(x-a) _____________________________________ b) V = 0 at positions x/a = -1,0.333 , on the -axis _______________________________ c)when x >>a the answer to part A becomes V = -kx /a ______________________________________… At large 'x' , the potential due to -2q is nearly twice the potential due to +q , hence resultant potrential is due to sum of - 2q and +q which is - q Potential at 'x' = - kx/a

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