Three point charges sitting on the x-axis. The positive x-direction is to the ri

Question

Three point charges sitting on the x-axis. The positive x-direction is to the right. The three charges have the following values for their charge and position: . Q1 =-6.1x10-6 C, x,--0.017 m Q2-5x1o% C, x2 0.059 m ·Q3 = 4.4x10-6 C, x3 = 0 m What are the magnitude and direction of the total electrostatic force on Q2? For the magnitude, give your answer in Newtons to 3 significant digits. (However, do NOT include the units in your answer.) The sign of your answer will give the direction: a positive answer implies the force is to the right and a negative answer implies the force is to the left.o

Explanation / Answer

From the given cordinates of the charges, it is clear that charge Q2 is at the extreme right and Q3 is at the middle of the charges.

Now, force on Q2 due to Q3 -

F1 = k*Q2*Q3 / d1^2 = (9 x 10^9 x 5 x 10^-6 x 4.4 x 10^-6) / (0.059)^2

= 56.88 N

Q2 is negative charged and Q3 is positive charged so there will a attractive force between the two charges.

So, direction of F1 will be in the left.

Means, F1 = -56.88 N

Now, force on Q2 due to Q1 charge -

F2 = k*Q2*Q1 / d2^2 = (9 x 10^9 x 5 x 10^-6 x 6.1 x 10^-6) / (0.059 + 0.017)^2

= 47.52 N

Both the charges Q2 and Q1 are negatively charged. So repulsive force will act between them.

Therefore, the net force on Q2 due to Q1 and Q3 -

F = F1 + F2 = -56.88 N + 47.52 N = -9.36 N

Negative sign shows that the net force is towards in the left direction.

So, your answer = -9.36

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