We need to find the point along the line joining two charges where the electric field is zero. The charges are \( q_1 = -3.30 \, \mu \mathrm{C} \) and \( q_2 = 7.00 \, \mu \mathrm{C} \), and they are 1.00 m apart. The electric field due to a point charge is given by \( E = \frac{k|q|}{r^2} \), where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.
Since \( q_1 \) is negative and \( q_2 \) is positive, the point where the electric field is zero must be outside the segment joining the charges. It will be closer to the smaller magnitude charge, \( q_1 \), because the electric field strength decreases with distance.
Assume the point is at a distance \( x \) to the left of \( q_1 \). The distance from \( q_2 \) to this point is \( x + 1.00 \, \mathrm{m} \). The electric field due to \( q_1 \) at this point is \( E_1 = \frac{k|q_1|}{x^2} \) and due to \( q_2 \) is \( E_2 = \frac{k|q_2|}{(x + 1.00)^2} \).
For the electric field to be zero, these must be equal in magnitude:
\[
\frac{k|q_1|}{x^2} = \frac{k|q_2|}{(x + 1.00)^2}
\]
Cancel \( k \) from both sides and solve for \( x \):
\[
\frac{|q_1|}{x^2} = \frac{|q_2|}{(x + 1.00)^2}
\]
\[
\frac{3.30}{x^2} = \frac{7.00}{(x + 1.00)^2}
\]
Cross-multiply to solve for \( x \):
\[
3.30(x + 1.00)^2 = 7.00x^2
\]
Expanding and simplifying:
\[
3.30(x^2 + 2x + 1) = 7.00x^2
\]
\[
3.30x^2 + 6.60x + 3.30 = 7.00x^2
\]
\[
0 = 7.00x^2 - 3.30x^2 - 6.60x - 3.30
\]
\[
0 = 3.70x^2 - 6.60x - 3.30
\]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3.70 \), \( b = -6.60 \), and \( c = -3.30 \):
\[
x = \frac{-(-6.60) \pm \sqrt{(-6.60)^2 - 4 \cdot 3.70 \cdot (-3.30)}}{2 \cdot 3.70}
\]
\[
x = \frac{6.60 \pm \sqrt{43.56 + 48.84}}{7.40}
\]
\[
x = \frac{6.60 \pm \sqrt{92.40}}{7.40}
\]
\[
x = \frac{6.60 \pm 9.615}{7.40}
\]
The positive root is:
\[
x = \frac{6.60 + 9.615}{7.40} \approx 2.1878
\]
The point where the electric field is zero is approximately 2.1878 m to the left of \( q_1 \).
\[
\boxed{x = 2.1878 \, \text{m to the left of } q_1}
\]
Answered
