2)Two charges attract each other with a force of 2.6 N. What will be the force i

Question

2)Two charges attract each other with a force of 2.6 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

3)wo point charges are fixed on the y axis: a negative point charge q1 = -31 ?C at y1 = +0.18 m and a positive point charge q2 at y2 = +0.34 m. A third point charge q = +9.0 ?C is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 21 N and points in the +y direction. Determine the magnitude of q2.

4)A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.56 m. A third charge of +2q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

5)

6)

At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 288 N/C. At a distance r2 from the charge, the field has a magnitude of 169 N/C. Find the ratio r2/r1.

8)A 6.8 ?C point charge is placed in an external uniform electric field that has a magnitude of 33000 N/C. At what distance from the charge is the net electric field zero?

10)

Explanation / Answer

1. F = k × q × q / r²
where
F = force between the particles = 2.9 N
k = constant of coulomb = 9×10^9 Nm²/C²
q = charge on particle 1 = 2.8×10^-6 C
q = charge on particle 2 = ?
r = distance between the particles = 0.28 m
so
2.9 = 9×10^9 × 2.8×10^-6 × q / 0.28²
q = (2.9 × 0.28²) / (9×10^9 × 2.8×10^-6)
q = 9.02×10^-6
q = 9,02 µC

So far, so good, but...
It is an ATTRACTIVE force, so the charges must have DIFFERENT SIGNS.
q is positive, so q must be negative.

q = -9.02 µC

2.solution

By F = q1q2/r^2
=>2.6 = q1q2/r^2 ---------------(i)
2nd case:-
=>F = q1q2/(r/9)^2
=>F = 81 x q1q2/r^2
=>F = 81 x 2,6
=>F = 210.6 N

3.

The attractive force (+y direction) on q due to q1 is
F1 = k*(31)(9)*10^-12 / 0.18^2 = 77.5 N ( here k = 9.0*10^9)

So the repulsive force (-y direction) on q due to q2 is
F2 = 77.5 - 21.0 N = 56.5 N
So
56.5 = k*(q2)(9)*10^-6 / 0.34^2
q2 = 8.06*10^-5 C = 80.06 C

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