AP Physics 2 Exam Question

Two metal plates, A and B, are placed parallel to each other, with a separation distance of 0.03 meters. Plate A has a charge of +8 microcoulombs, while plate B has a charge of -12 microcoulombs.

a) Calculate the electric field magnitude between the plates. b) Determine the direction of the electric field between the plates. c) If a positive test charge of 2 microcoulombs is placed at a point between the plates and experiences an electric force of magnitude 0.5 Newton, calculate the electric field magnitude at that point. d) Is the electric field at this point directed towards plate A or plate B? e) If the separation distance between the plates is doubled, will the electric field magnitude between the plates increase, decrease, or remain the same?

Answer

a) The electric field magnitude between the plates can be calculated using the formula:

Where:

• E is the electric field magnitude
• Q is the charge on either plate (assumed to be the same)
• $\varepsilon_0$ is the vacuum permittivity constant ($8.85 \times 10^{-12} \, \text{{C}}^2 / \text{{Nm}}^2$)
• d is the separation distance between the plates

Using the given values, we have: [ E = \frac{{|Q|}}{{4 \pi \varepsilon_0 d}} = \frac{{(8 \times 10^{-6})}}{{4 \pi (8.85 \times 10^{-12}) (0.03)}} = \frac{{8 \times 10^{-6}}}{{4 \pi (8.85 \times 10^{-12} \times 0.03)}} , \text{{N/C}} ]

To calculate this value, we can use a calculator and the value of π to get: [ E = \frac{{8 \times 10^{-6}}}{{4 \times 3.141592 \times (8.85 \times 10^{-12} \times 0.03)}} , \text{{N/C}} ]

Calculating this value gives us approximately 2699485 N/C.

b) The electric field between the plates is directed from positive charge to negative charge. So in this case, the electric field is directed from plate A towards plate B.

c) The electric field magnitude at a point can be calculated using the formula:

Where:

• E is the electric field magnitude at the point
• F is the electric force magnitude experienced by the test charge
• q is the magnitude of the test charge

Using the given values, we have: [ E = \frac{{F}}{{|q|}} = \frac{{0.5}}{{2 \times 10^{-6}}} , \text{{N/C}} ]

Simplifying this expression gives us 250000 N/C.

d) Since the test charge is positive, its direction of motion would be from positive to negative charges. Therefore, the electric field at this point is directed towards plate B.

e) When the separation distance between the plates is doubled, the electric field magnitude between the plates would decrease. This can be observed from the formula for electric field magnitude:

As the separation distance (d) increases, the denominator of the equation also increases, resulting in a smaller electric field magnitude (E).