Post

Electric Field and Forces Between Parallel Plates

Created by @nathanedwards
 at October 31st 2023, 5:15:19 pm.
AP_Physics_2-Questions
#electric-field
#parallel-plates
#force-on-charge

Question:

Two parallel plates, labeled A and B, are placed a distance of 0.02 meters apart. Plate A carries a charge of +5.0 microcoulombs, while Plate B carries a charge of -3.0 microcoulombs.

  1. Calculate the magnitude of the electric field between the plates.
  2. If a positive charge of 1.6 microcoulombs is placed between the plates at a point P, what is the direction and magnitude of the force experienced by this charge? Explain your answer in detail.

Answer:

1. Calculation of Electric Field:

The electric field between two parallel plates can be calculated using the formula:

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

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

Given: Potential difference, V = ? Distance between the plates, d = 0.02 m

To find the potential difference, we can make use of the formula:

V=Ed V = Ed

Substituting the given values, we have:

V=E×0.02 V = E \times 0.02

Now, to find the electric field, we can use the formula:

E=Fq E = \frac{F}{q}

where E is the electric field, F is the force experienced by the charge, and q is the charge.

From Coulomb's Law, we know that the force experienced by the charge is given by:

F=k⋅q1⋅q2r2 F = \frac{k \cdot q_1 \cdot q_2}{r^2}

where k is the Coulomb's constant (9 x 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between them.

Given: Charge, q1 = +5.0 μC = +5.0 x 10^-6 C Charge, q2 = -3.0 μC = -3.0 x 10^-6 C Distance, r = 0.02 m

Substituting the given values and solving for F:

F=(9×109)⋅(5.0×10−6)⋅(3.0×10−6)(0.02)2 F = \frac{(9 \times 10^9) \cdot (5.0 \times 10^-6) \cdot (3.0 \times 10^-6)}{(0.02)^2}

Simplifying the expression:

F=9×5.0×3.0×10−6(0.02)2×109 F = \frac{9 \times 5.0 \times 3.0 \times 10^{-6}}{(0.02)^2} \times 10^9
F=675×10−6×109 F = 675 \times 10^{-6} \times 10^9
F=675×103 F = 675 \times 10^3
F=675,000N F = 675,000 N

Now we can find the electric field using the formula:

E=Fq E = \frac{F}{q}

Given: Force, F = 675,000 N Charge, q = 1.6 μC = 1.6 x 10^-6 C

Substituting the given values and solving for E:

E=675,0001.6×10−6 E = \frac{675,000}{1.6 \times 10^-6}

Simplifying the expression:

E=675,0001.6×106 E = \frac{675,000}{1.6} \times 10^6
E=421,875×106 E = 421,875 \times 10^6
E=4.21875×108N/C E = 4.21875 \times 10^8 N/C

Therefore, the magnitude of the electric field between the plates is 4.21875 x 10^8 N/C.

2. Calculation of Force on a Charge:

The force experienced by a charge in an electric field 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.

Given: Charge, q = 1.6 μC = 1.6 x 10^-6 C Electric field, E = 4.21875 x 10^8 N/C

Substituting the given values and solving for F:

F=(1.6×10−6)×(4.21875×108) F = (1.6 \times 10^-6) \times (4.21875 \times 10^8)

Simplifying the expression:

F=6.749×102 F = 6.749 \times 10^2
F=674.9N F = 674.9 N

Since the charge is positive, it experiences a force in the opposite direction of the electric field. Therefore, the force experienced by the charge at point P is 674.9 N in the opposite direction of the electric field.

Chat

Save To Bookmark