Question

A charge +q is at the origin. A charge-2q is at x = 8.90 m on the +x axis. (a) For what finite value of x is the electric field zero? 3.69 Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) For what finite values of x is the electric potential zero? (Note: Assume a reference level of potential V Smallest value of x 0 at roo.) Largest value of x:

Answer 1

(a)

formula for electric field due to point charge is

E = kq/ r^2

let P be the point from the +q charge where otal electric field is zero

E total= E1 + E2

kq/ x^2 + k ( -2q)/ ( x- 8.90)^2= 0

kq/ x^2 = k ( 2q)/ ( x- 8.90)^2

( x-8.90)^2/ x^2 = 2

( x-8.90)^2 = 2 x^2

x^2 -17.8 x+79.21 = 2x^2

x^2 +17.8 x -79.21 =0

solving quadratic equation

x= -21.48 m or 3.68 m

in between both charges electric field never zero so x = -21.48 m

(b)

formula for potential due to point charge is

V = kq/r

V net = kq/ x- k2q/ ( 8.9-x)

0 =  kq/ x- k2q/ ( 8.9-x)

kq/ x= k2q/ ( 8.9-x)

8.9-x = 2x

8.9= 3x

x= 2.96 m ( largest)

for smallest

kq/ x=- k2q/ ( 8.9-x)

8.9-x =- 2x

8.9= -x

x=-8.9 m is smallest value