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Electric Field Calculation for Two Point Charges

Created by @nathanedwards
 at November 4th 2023, 9:51:44 pm.
AP_Physics_2-Questions
#electric-field
#point-charges
#electrostatics

Electric Fields

Question:

Two point charges are placed on the x-axis. Charge A has a value of +3 μC and is located at x = -2 m, while charge B has a value of -6 μC and is located at x = +4 m. Determine the electric field at a location on the x-axis where x = +1 m.

Answer:

To determine the electric field at x = +1 m, we need to calculate the electric field contribution from each charge and then sum them up.

The electric field due to a point charge can be calculated using the formula:

E=k⋅∣q∣r2 E = \frac{{k \cdot |q|}}{{r^2}}

Where:

  • E is the electric field
  • k is the electrostatic constant, approximately equal to 9.0 x 10^9 Nm²/C²
  • |q| is the magnitude of the charge
  • r is the distance from the charge to the point where the electric field is being calculated

First, let's calculate the electric field due to charge A at x = +1 m.

q_A = +3 μC (positive because it is a positive charge) r_A = x - x_A = 1 m - (-2 m) = 3 m (taking the positive direction since the point is to the right of the charge)

Using the electric field formula, we can calculate the electric field due to charge A:

EA=k⋅∣qA∣rA2 E_A = \frac{{k \cdot |q_A|}}{{r_A^2}}
EA=9.0×109 Nm²/C²⋅3×10−6 C(3 m)2 E_A = \frac{{9.0 \times 10^9 \,\text{Nm²/C²} \cdot 3 \times 10^{-6} \,\text{C}}}{{(3 \,\text{m})^2}}
EA=27.0×103 Nm²/C9 m² E_A = \frac{{27.0 \times 10^3 \,\text{Nm²/C}}}{{9 \,\text{m²}}}
EA=3.0×103 N/C E_A = 3.0 \times 10^3 \,\text{N/C}

Next, let's calculate the electric field due to charge B at x = +1 m.

q_B = -6 μC (negative because it is a negative charge) r_B = x - x_B = 1 m - 4 m = -3 m (taking the negative direction since the point is to the left of the charge)

Using the electric field formula, we can calculate the electric field due to charge B:

EB=k⋅∣qB∣rB2 E_B = \frac{{k \cdot |q_B|}}{{r_B^2}}
EB=9.0×109 Nm²/C²⋅6×10−6 C(−3 m)2 E_B = \frac{{9.0 \times 10^9 \,\text{Nm²/C²} \cdot 6 \times 10^{-6} \,\text{C}}}{{(-3 \,\text{m})^2}}
EB=54.0×109 Nm²/C9 m² E_B = \frac{{54.0 \times 10^9 \,\text{Nm²/C}}}{{9 \,\text{m²}}}
EB=6.0×109 N/C E_B = 6.0 \times 10^9 \,\text{N/C}

Now, let's calculate the total electric field at x = +1 m by summing up the electric field contributions from charge A and charge B:

E_total = E_A + E_B

E_total = (3.0 x 10^3 N/C) + (6.0 x 10^9 N/C)

E_total = 6.0 x 10^9 N/C

Therefore, the electric field at x = +1 m is 6.0 x 10^9 N/C (in the positive x-direction).

Note: In this problem, we assumed that the point charges are so small compared to the distance between them that we can treat them as point charges. If the distance between the charges was not small compared to the size of the charges, we would need to consider the finite size of the charges and their distribution of charge.

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