Question

A scientist fixes a test charge q′ to point P and then measures the electrostatic force it experiences there. She then calculates the magnitude of the electric field by applying the equation E = F'/q' . Which change to her procedure would vary the value of E that she calculates?
A. keep the test charge in the same location, but alter its charge from q′ to −q'
B. keep the charge on the test charge the same, but move the test charge to a different position fa
C .keep the test charge in the same location, but alter its charge from q′ to 2q

Answer 1

Answer:

Correct answer: B.

Explanation:

Electrostatic Field

It measures the electric effect of a charge distribution in its surroundings. If we wanted to test or measure the electric field by using a point-charge of value q', then the electric field is

$\displaystyle E=\frac{F}{q'}$

The electrostatic force between two point charges is

$\displaystyle F=\frac{k\ q_1\ q'}{r^2}$

So, the electric field is

$\displaystyle E=\frac{\frac{k\ q_1\ q'}{r^2}}{q'}$

$\displaystyle E=\frac{k\ q_1}{r^2}$

The theory shows that the electric field doesn't depend on the test charge used, i.e., if we now use q'=-q', the electric force will be of the same magnitude but in the opposite direction, thus the electric field will be the same. Let's recall the formula is used to compute the scalar value of the field, the direction must be studied separately.

Now, if we changed the test charge to another value, say 2q, the measured force will be

$\displaystyle F=\frac{k\ q_1\ 2q}{r^2}$

And the new electric field is

$\displaystyle E=\frac{k\ q_1}{r^2}$

We can see the electric field is not affected by the value of the test charge.

Finally, if we move the test charge to another location and keep the same charge, the electric force will vary and the electric field will be different.

Correct answer: B.