Question

Problem 5: A point charge of 54 iC is placed at the origin (x1 0) of a coordinate system, and another charge of-1.3 μ0is placed placed on the x-a us at X-t. 29 m. wh e n fiel d strength is tro、10cc c ; s zefO Part (a) Where on the r-axis can a third charge be placed in meters so that the net force on it is zero? Numeric A numeric value is expected and not an expression Part (b) What if both charges are positive, that is, what if the second charge is 1.3 C? Numeric A numeric value is expected and not an expression.

how would I work out part a &b

Answer 1

a)

force will be zero at that point where field strength is zero.

field at point x = [ k(5.4*10^-6)/(x)² ] - [ k (1.3*10^-6)/(x - 0.29)² ] = 0

5.4 / x² = 1.3/(x - 0.29)²

5.4*x² - 3.13*x + 0.45 = 1.3*x²

4.1*x² - 3.13*x + 0.45 = 0

x = 0.57 or 0.19 m

and that point will be beyond 0.29 m

so x = 0.57 m

b)

now,

point will be in between charges.

so x = 0.19 m