Question

what is the magnitude of the electric field at a point midway between a −8.0 μc and a +6.6 μc charge 8.2 cm apart? assume no other charges are nearby express your answer using two significant figures. e = value units submit request answer part b what is the direction of the electric field? ○ toward the negative charge ○ toward the positive charge submit request answer

Explanation:

Part A

Step1: Find the distance to the midpoint

The total distance between the charges is \( d = 8.2\ \text{cm} = 0.082\ \text{m} \). The distance to the midpoint from each charge is \( r=\frac{d}{2}=\frac{0.082}{2}=0.041\ \text{m} \).

Step2: Recall the electric field formula

The electric field due to a point charge is given by \( E = k\frac{|q|}{r^{2}} \), where \( k = 8.988\times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 \). For the two charges, the electric fields at the midpoint will add up (since they are in the same direction: the field from the negative charge points towards it, and the field from the positive charge also points towards the negative charge, or away from the positive charge, so they are in the same direction).
Let \( q_1=- 8.0\ \mu\text{C}=-8.0\times 10^{- 6}\ \text{C} \) and \( q_2 = 6.6\ \mu\text{C}=6.6\times 10^{-6}\ \text{C} \)
The magnitude of the electric field due to \( q_1 \) is \( E_1=k\frac{|q_1|}{r^{2}} \) and due to \( q_2 \) is \( E_2=k\frac{|q_2|}{r^{2}} \)
The total electric field \( E = E_1+E_2=k\frac{|q_1| + |q_2|}{r^{2}} \)

Step3: Substitute the values

First, calculate \( |q_1|+|q_2|=(8.0\times 10^{-6}+6.6\times 10^{-6})\ \text{C}=14.6\times 10^{-6}\ \text{C} \)
\( r = 0.041\ \text{m} \), so \( r^{2}=(0.041)^{2}=0.001681\ \text{m}^2 \)
\( E=8.988\times 10^{9}\times\frac{14.6\times 10^{-6}}{0.001681} \)
\( E=8.988\times 10^{9}\times8.685\times 10^{-3} \)
\( E\approx7.8\times 10^{7}\ \text{N/C} \) (after rounding to two significant figures)

The electric field due to a negative charge points towards the negative charge, and the electric field due to a positive charge points away from the positive charge (towards the negative charge in this case, since the positive charge is on the other side of the midpoint from the negative charge). So the net electric field direction is toward the negative charge.

Answer:

\( E = \boxed{7.8\times 10^{7}\ \text{N/C}} \)

Part B