Answer:
The intensity of the net electric field will:
[tex]E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C[/tex]
Explanation:
Here we need first find the electric field due to the first charge at the midway point.
The electric field equation is given by:
[tex]|E_{1}|=k\frac{q_{1}}{d^{2}}[/tex]
Where:
So, we will have:
[tex]|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}[/tex]
[tex]|E_{1}|=1.8*10^{7}\: N/C[/tex]
The direction of E1 is to the right of the midpoint.
Now, the second electric field is:
[tex]|E_{2}|=k\frac{q_{2}}{d^{2}}[/tex]
[tex]|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}[/tex]
[tex]|E_{2}|=7.2*10^{6}\: N/C[/tex]
The direction of E2 is to the right of the midpoint because the second charge is negative.
Finally, the intensity of the net electric field will:
[tex]E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C[/tex]
I hope it helps you!
The net electric field midway between these two charges is 2.52 x 10⁷ N/C.
The electric field strength between two point charges is the force per unit charge.
Electric field strength due to +20.0 μC charge
[tex]E _1 = \frac{kq_1}{r^2} \\\\E_ 1 = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{(0.1)^2} \\\\E _1 = 18 \times 10^6 \ N/C[/tex]
Electric field strength due to -8μC charge
[tex]E _2 = \frac{kq_2}{r^2} \\\\E_ 2 = \frac{9 \times 10^9 \times 8 \times 10^{-6}}{(0.1)^2} \\\\E _2 = 7.2 \times 10^6 \ N/C[/tex]
The net electric field midway between these two charges is calculated as follows;
[tex]E_{net} = 18 \times 10^6 \ + \ 7.2 \times 10^6\\\\E_{net} = 25.2 \times 10^6 \ N/C\\\\E_{net} = 2.52 \times 10^7 \ N/C[/tex]
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