A charge of 7.00 multiplied by 10-9 C and a charge of -2.50 multiplied by 10-9 C

Question

A charge of 7.00 multiplied by 10-9 C and a charge of -2.50 multiplied by 10-9 C are separated by a distance of 60.0 cm. Find the position at which a third charge, of 14.0 multiplied by 10-9 C, can be placed so that the net electrostatic force on it is zero
......... m (beyond the -2.50 multiplied by 10-9 C charge)

Explanation / Answer

q1 = 7E-9 C, q2 = -2.5E-9 C, x1 = 0m, x2 = 0.6 m R = x2-x1 = 0.6 m S = sign(q1q2) = -1 d1 = R/(1+S*sqrt(|q2/q1|)) = 2.5398181667894 m d2 = R-d1 = -1.9398181667894 m [Derivation of d1 equation: E = k(q1/d1^2+q2/d2^2) = 0 ==> d2/d1 = -sqrt(|q2/q1|) ==> R/d1 = 1-sqrt(|q2/q1|) ==> d1 = R/(1-sqrt(|q2/q1|))] Null-field point (point of null net force on a 3rd charge) is at -d2 = 1.9398181667894 m beyond q2 at x2 (

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