Question

determine the electric field strength at point p in the following situation where charge a is -4.00 μc and charge b is -3.00 μc. 6.89 x 10⁶ n/c to the left a p b -3.00×10⁻⁶c 5.00 cm 6.00 cm

Explanation:

Step1: Recall Electric Field Formula

The electric field due to a point charge is given by \( E = k\frac{|q|}{r^2} \), where \( k = 8.988\times10^{9}\, \text{N·m}^2/\text{C}^2 \), \( q \) is the charge, and \( r \) is the distance from the charge to the point.

Step2: Calculate \( E_A \) (Field from Charge A)

Charge \( A \): \( q_A = -4.00\,\mu\text{C} = -4.00\times10^{-6}\,\text{C} \), distance \( r_A = 5.00\,\text{cm} = 0.0500\,\text{m} \).
\( E_A = k\frac{|q_A|}{r_A^2} = 8.988\times10^{9} \times \frac{4.00\times10^{-6}}{(0.0500)^2} \).
Calculate numerator: \( 8.988\times10^{9} \times 4.00\times10^{-6} = 35952 \).
Denominator: \( (0.0500)^2 = 0.0025 \).
\( E_A = \frac{35952}{0.0025} = 1.43808\times10^{7}\,\text{N/C} \).
Direction: Since \( A \) is negative, field points toward \( A \) (to the left).

Step3: Calculate \( E_B \) (Field from Charge B)

Charge \( B \): \( q_B = -3.00\,\mu\text{C} = -3.00\times10^{-6}\,\text{C} \), distance \( r_B = 6.00\,\text{cm} = 0.0600\,\text{m} \).
\( E_B = k\frac{|q_B|}{r_B^2} = 8.988\times10^{9} \times \frac{3.00\times10^{-6}}{(0.0600)^2} \).
Numerator: \( 8.988\times10^{9} \times 3.00\times10^{-6} = 26964 \).
Denominator: \( (0.0600)^2 = 0.0036 \).
\( E_B = \frac{26964}{0.0036} = 7.49\times10^{6}\,\text{N/C} \).
Direction: Since \( B \) is negative, field points toward \( B \) (to the right).

Step4: Find Net Electric Field

Net field \( E_{\text{net}} = E_A - E_B \) (since \( E_A \) is left, \( E_B \) is right; left is positive here).
\( E_{\text{net}} = 1.43808\times10^{7} - 7.49\times10^{6} \).
Convert to same exponent: \( 14.3808\times10^{6} - 7.49\times10^{6} = 6.89\times10^{6}\,\text{N/C} \) (to the left, as \( E_A > E_B \)).

Answer:

\( 6.89\times10^{6}\,\text{N/C} \) to the left