Question

what is the magnitude of the electric field at a point midway between a - 5.3 μc and a +8.2 μc charge 9.8 cm apart? assume no other charges are nearby. express your answer using two significant figures.
e = value units
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part b
what is the direction of the electric field?
toward the negative charge
toward the positive charge
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Explanation:

Step1: Calculate the distance to the mid - point

The two charges are $d = 9.8\ cm=0.098\ m$ apart. The distance from each charge to the mid - point is $r=\frac{d}{2}=\frac{0.098}{2}=0.049\ m$.

Step2: Recall the electric - field formula

The electric - field due to a point charge is $E = k\frac{|q|}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$.

Step3: Calculate the electric field due to each charge

For the charge $q_1=- 5.3\ \mu C=-5.3\times10^{-6}\ C$, the magnitude of the electric field it creates at the mid - point is $E_1=k\frac{|q_1|}{r^{2}}=(9\times10^{9})\frac{5.3\times10^{-6}}{(0.049)^{2}}$.
For the charge $q_2 = + 8.2\ \mu C=8.2\times10^{-6}\ C$, the magnitude of the electric field it creates at the mid - point is $E_2=k\frac{|q_2|}{r^{2}}=(9\times10^{9})\frac{8.2\times10^{-6}}{(0.049)^{2}}$.

Step4: Calculate the net electric field

Since the electric fields due to the two charges are in the same direction (because one charge is positive and the other is negative), the net electric field $E = E_1 + E_2$.
\[

$$\begin{align*} E&=(9\times10^{9})\frac{5.3\times10^{-6}+8.2\times10^{-6}}{(0.049)^{2}}\\ &=(9\times10^{9})\frac{13.5\times10^{-6}}{(0.049)^{2}}\\ &=(9\times10^{9})\frac{13.5\times10^{-6}}{0.002401}\\ &=\frac{121.5\times10^{3}}{0.002401}\\ &\approx5.1\times10^{7}\ N/C \end{align*}$$

\]

Step5: Determine the direction of the electric field

The electric field points from the positive charge to the negative charge. So the direction is toward the negative charge.

Answer:

$E = 5.1\times10^{7}\ N/C$
toward the negative charge