Two point charges, Q 1 = 3.3 ? C and Q 2 = -1.5 ? C ,are placed on the x axis. S

Question

Two point charges, Q1 = 3.3?C and Q2 = -1.5?C ,are placed on the x axis. Suppose that Q2 is placed at the origin, and Q1 is placed at the coordinatex1 = ? 6.0cm (Figure 1) .

Part A:

At what point(s) along the x axis is the electric field zero? Determine the x-coordinate(s) of the point(s).

Express your answer using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas.

Part B:

At what point(s) along the x axis is the potential zero? Determine the x-coordinate(s) of the point(s).

Express your answer using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas.

Explanation / Answer

Part A)

The formula for the Electric Field is E = kq/r2 and we need the two E fields to cancel. This location will be somewhere along the negative x axis. It needs to be closer to the smaller charge and farther from the larger charge to cancel.

So kq/r2 = kq/r2  (The k's will cancel since they are the same)

1.5/x2 = 3.3/(x+6)2  -- Cross Multiply

1.5(x + 6)2 = 3.3x2 -- Distribute and expand the binomial

1.5x2 + 18x + 54 = 3.3x2  -- Put in standard form

1.8x2 - 18x - 54 = 0

Put into the quadratic equation to find the roots.

The roots are...12.4 and -2.42. We can eliminate the negative root since that puts the distance between the charges where the field adds, and does not cancel.

Thus the distance is 12.4 cm along the negative x axis.

That is at location -12 cm (Using two sig figs)

Part B

This is similar, but the formula is different, and there can be two locations, for the first...

V = kq/r, so...

q/r = q/r

1.5/x = 3.3/(x + 6)

1.5x + 9 = 3.3x

9 = 1.8x

x = 5 cm

So one location is at -5.0 cm on the x axis.

For the other

1.5/x = 3.3/6-x

9 - 1.5x = 3.3x

9 = 4.8x

x = 1.9 cm

The other location is +1.9 cm on the positive x axis

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