Question

6 18.3.P016 0/20 Submissions Used A test charge of +7 JC is placed halfway between a charge of +6 JC and another of +8 C separated by 18 cm (a) What is the magnitude of the force (in N) on the test charge? IN (b) What is the direction of this force (away from or toward the +6 HC charge)? O away from the +6 JC charge O toward the +6 JC charge Additional Materials Reading Submit Answer Save Progress

Answer 1

a)

net force acting on 7 uC is vector sum of individual forces due to 6uC and 8 uC.

force due to 6 uC is repulsive hence directed to the 8 uC

Force due to 8 uC is also repulsive on 7 uC hence directed to the 6uC

Now Net force is given by

F = F₈ - F₆​​​​​​

F = k q q₈ / r^2 - k q q₆/ r^2

F = 9*10^9* 7*10^-6*8*10^-6 / 0.09^2 - 9*10^9* 7*10^-6* 6*10^-6/ 0.09^2

F = 15.556 N

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b)

Since force between act due to 8uC is greater than 6 uC so net force is towards 6 uC.

option (b) is correct

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