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Electric Field Between Parallel Plates and Charge Interaction

Created by @nathanedwards
 at October 31st 2023, 9:37:20 pm.
AP_Physics_2-Questions
#physics
#electric-fields
#parallel-plates

Question:

Two parallel metal plates, labeled A and B, are 1.5 cm apart. Plate A has a charge of +12 nC and Plate B has a charge of -8 nC.

a) Calculate the electric field intensity between the plates.

b) A negatively charged object with a charge of -2 nC is placed at a point 0.5 cm above Plate B. Calculate the electric field intensity at that point due to the combination of the plates and the object.

c) If the negatively charged object is released, calculate the direction and magnitude of the resulting force on the object.

Assume the electric permittivity of free space, ε₀, is 8.85 x 10⁻¹² C²/N∙m².

Answer:

a) To calculate the electric field intensity between the plates, we can use the formula for the electric field between parallel plates:

E=Vd E = \frac{{V}}{d}

where E is the electric field intensity, V is the voltage difference between the plates, and d is the distance between the plates.

First, we need to calculate the voltage difference between the plates. Since the plates have opposite charges, the voltage difference can be found using the formula:

V=QC V = \frac{{Q}}{{C}}

where Q is the absolute value of the charge and C is the capacitance of the system. In this case, plate A has a charge of +12 nC and plate B has a charge of -8 nC, so the total charge is 20 nC.

The capacitance of the system is given by the formula:

C=QV C = \frac{{Q}}{{V}}

where Q is the absolute value of the charge and V is the voltage difference. In this case, Q is 20 nC and V is unknown. Rearranging the formula, we can solve for V:

V=QC=20×10−9 CC V = \frac{{Q}}{{C}} = \frac{{20 \times 10^{-9}\,\text{C}}}{{C}}

Substituting this equation into the formula for electric field intensity, we get:

E=Vd=20×10−9 CC×1.5×10−2 m E = \frac{{V}}{{d}} = \frac{{20 \times 10^{-9}\,\text{C}}}{{C \times 1.5 \times 10^{-2}\,\text{m}}}

Now, we need to find the capacitance of the system. The capacitance of a parallel plate capacitor can be calculated using the formula:

C=ε0Ad C = \frac{{\varepsilon₀ A}}{{d}}

where ε₀ is the electric permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this case, A is unknown, so we can assume the plates have a very large area compared to their separation, resulting in a uniform electric field between them. Assuming A is very large, we can ignore it in our calculations and simply use the value of ε₀:

C=ε0d C = \frac{{\varepsilon₀}}{{d}}

Plugging in the values, we get:

C=8.85×10−12 C²/N∙m²1.5×10−2 m C = \frac{{8.85 \times 10^{-12}\,\text{C²/N∙m²}}}{{1.5 \times 10^{-2}\,\text{m}}}

Plugging this value of C back into the formula for electric field intensity, we can calculate E:

E=20×10−9 C(8.85×10−12 C²/N∙m²)×(1.5×10−2 m) E = \frac{{20 \times 10^{-9}\,\text{C}}}{{(8.85 \times 10^{-12}\,\text{C²/N∙m²}) \times (1.5 \times 10^{-2}\,\text{m})}}

Calculating this expression, we find:

E≈8926 N/C E ≈ 8926 \, \text{N/C}

Therefore, the electric field intensity between the plates is approximately 8926 N/C.

b) To calculate the electric field intensity at a point 0.5 cm above Plate B, we can use the formula for the electric field due to a point charge:

E=k⋅Qr2 E = \frac{{k \cdot Q}}{{r^2}}

where E is the electric field intensity, k is the Coulomb's constant (k = 8.99 x 10^9 N·m^2/C^2), Q is the charge, and r is the distance from the charge.

In this case, the distance from the negatively charged object to the desired point is 0.5 cm (or 0.005 m) and the charge is -2 nC. Plugging these values into the formula, we get:

E=(8.99×109 N\cdotpm²/C²)⋅(−2×10−9 C)(0.005 m)2 E = \frac{{(8.99 \times 10^9\,\text{N·m²/C²}) \cdot (-2 \times 10^{-9}\,\text{C})}}{{(0.005\,\text{m})^2}}

Performing the calculation, we find:

E≈−7192000 N/C E ≈ -7192000 \, \text{N/C}

Therefore, the electric field intensity at the point 0.5 cm above Plate B is approximately -7192000 N/C. The negative sign indicates that the field points downward.

c) When the negatively charged object is released, it will experience a force due to the electric field created by the combined plates and the object itself. The force can be calculated using the formula:

F=Q⋅E F = Q \cdot E

where F is the force, Q is the charge, and E is the electric field intensity.

In this case, the charge of the object is -2 nC and the electric field intensity at its position is -7192000 N/C (downward). Plugging these values into the formula, we get:

F=(−2×10−9 C)⋅(−7192000 N/C) F = (-2 \times 10^{-9}\,\text{C}) \cdot (-7192000\,\text{N/C})

Simplifying the expression, we find:

F≈1.44×10−2 N F ≈ 1.44 \times 10^{-2}\,\text{N}

Therefore, the resulting force on the negatively charged object is approximately 1.44 x 10^-2 N, acting in the downward direction.

This concludes the answer to all parts of the question.

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