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Electric Field and Force: Charged Spheres Interaction

Created by @nathanedwards
 at November 1st 2023, 5:37:57 pm.
AP_Physics_2-Questions
#electric-field
#force
#coulombs-law

Question:

Two small spheres, A and B, are placed 1 meter apart from each other. Sphere A has a charge of +3 microcoulombs and sphere B has a charge of -5 microcoulombs.

a) Calculate the magnitude and direction of the electric field at a point that is located 0.5 meters to the right of sphere A.

b) If a small test charge of +2 nanocoulombs is placed at the given point, what force will act on it due to the electric field created by the two spheres?

Answer:

a) To calculate the electric field at the given point due to the charges on sphere A and B, we can use the principle of superposition.

The electric field from a single charge is given by Coulomb's law:

E=14πϵ0qr2E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

Where:

  • E is the electric field
  • q is the charge
  • r is the distance from the charge
  • ε₀ is the permittivity of free space, equal to 8.85×10−12 C2/N⋅m28.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2

For sphere A:

  • q₁ = +3 μC = 3×10−6 C3 \times 10^{-6} \, \text{C}
  • r₁ = 0.5 meters to the right of sphere A

The electric field due to sphere A is:

E1=14πϵ0q1r12E₁ = \frac{1}{4\pi\epsilon_0}\frac{q₁}{r₁^2}

For sphere B:

  • q₂ = -5 μC = −5×10−6 C-5 \times 10^{-6} \, \text{C}
  • r₂ = 0.5 meters to the left of sphere B

The electric field due to sphere B is:

E2=14πϵ0q2r22E₂ = \frac{1}{4\pi\epsilon_0}\frac{q₂}{r₂^2}

Hence, the total electric field at the given point is the vector sum of the individual electric fields:

Etotal=E1+E2E_{\text{total}} = E₁ + E₂

Now, let's calculate the magnitude and direction of the electric field at the given point.

First, let's calculate the electric field due to sphere A:

E1=14πϵ0q1r12E₁ = \frac{1}{4\pi\epsilon_0}\frac{q₁}{r₁^2}
E1=14π(8.85×10−12 C2/N⋅m2)3×10−6 C(0.5 m)2E₁ = \frac{1}{4\pi(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)}\frac{3 \times 10^{-6} \, \text{C}}{(0.5 \, \text{m})^2}
E1=3.6×104 N/CE₁ = 3.6 \times 10^4 \, \text{N/C}

The electric field due to sphere A is positive and directed to the right.

Next, let's calculate the electric field due to sphere B:

E2=14πϵ0q2r22E₂ = \frac{1}{4\pi\epsilon_0}\frac{q₂}{r₂^2}
E2=14π(8.85×10−12 C2/N⋅m2)−5×10−6 C(0.5 m)2E₂ = \frac{1}{4\pi(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)}\frac{-5 \times 10^{-6} \, \text{C}}{(0.5 \, \text{m})^2}
E2=−9×104 N/CE₂ = -9 \times 10^4 \, \text{N/C}

The electric field due to sphere B is negative and directed to the left.

Hence, the total electric field at the given point is:

Etotal=E1+E2E_{\text{total}} = E₁ + E₂
Etotal=(3.6×104 N/C)+(−9×104 N/C)E_{\text{total}} = (3.6 \times 10^4 \, \text{N/C}) + (-9 \times 10^4 \, \text{N/C})
Etotal=−5.4×104 N/CE_{\text{total}} = -5.4 \times 10^4 \, \text{N/C}

The magnitude of the electric field at the given point is 5.4×104 N/C5.4 \times 10^4 \, \text{N/C} and it is directed to the left.

b) Now, let's calculate the force that will act on a +2 nanocoulomb test charge placed at the given point.

The force experienced by a charge in an electric field is given by:

F=q⋅EF = q \cdot E

Where:

  • F is the force
  • q is the charge
  • E is the electric field

In this case:

  • q = +2 nC = 2×10−9 C2 \times 10^{-9} \, \text{C}
  • E = -5.4 x 10^4 N/C

The force acting on the test charge is:

F=(2×10−9 C)⋅(−5.4×104 N/C)F = (2 \times 10^{-9} \, \text{C}) \cdot (-5.4 \times 10^4 \, \text{N/C})
F=−10.8×10−5 NF = -10.8 \times 10^{-5} \, \text{N}

The force acting on the test charge is approximately −1.08 μN-1.08 \, \mu\text{N}, directed to the left.

Therefore, the force acting on the test charge due to the electric field created by the two spheres is approximately −1.08 μN-1.08 \, \mu\text{N}, directed to the left.

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