Question

In: Physics

Two point charges, Q1 = 3.3?C and Q2 = -1.5?C ,are placed on the x axis....

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.

Solutions

Expert Solution

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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