A charge of 6.00 × 10–9 C and a charge of –3.00 × 10–9 C are...
A charge of 6.00 × 10–9 C and a charge of –3.00 × 10–9 C are separated by a distance of 60.0 cm. Find the position at which a third charge of 12.0 × 10–9 C can be placed so that the net electrostatic force on it is zero. ( x = 1.45 m beyond the –3.00 nC charge)
Can you please explain step by step and add a drawing for this too? The answers are in the brackets. Thanks so much!
Homework Answers
Solution) The electric force F is given by
F = Kq1q2/r^2
Here K = 9×10^(9) Nm^2/C^2
q1 and q2 are charges
r is separation distance
Here in this case charge q3 should be placed beyond q2 so in the figure q3 is placed to right of q2
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