Answer:
The net electric field is [tex]1.39\times10^{6}\ V/m[/tex]
Explanation:
Given that,
Charge at 0 mark = +5.0 μC
Charge at 50 mark = -4.0 μC
We need to calculate the electric field at 30 cm mark
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
Where, q = charge
r = distance
Put the value into the formula
[tex]E_{1}=\dfrac{8.99\times10^{9}\times5.0\times10^{-6}}{(30\times10^{-2})^2}[/tex]
[tex]E_{1}=499444.44\ V/m[/tex]
We need to calculate the electric field at 50 cm mark
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E_{2}=\dfrac{8.99\times10^{9}\times4.0\times10^{-6}}{(20\times10^{-2})^2}[/tex]
[tex]E_{2}=899000\ V/m[/tex]
We need to calculate the net electric field
Using formula of net electric field
[tex]E=E_{1}+E_{2}[/tex]
Put the value into the formula
[tex]E=499444.44+899000[/tex]
[tex]E=1398444.44\ V/m[/tex]
[tex]E=1.39\times10^{6}\ V/m[/tex]
Hence, The net electric field is [tex]1.39\times10^{6}\ V/m[/tex]
The net electric field will be "[tex]1.39\times 10^ 6[/tex] V/m".
According to the question,
0 mark charge = +5.0 μC
50 mark charge = -4.0 μC
We know the formula,
→ [tex]E = \frac{kq}{r^2}[/tex]
By substituting the values,
[tex]E_1 = \frac{8.99\times 10^9\times 5.0\times 10^{-6}}{(30\times 10^{-2})^2}[/tex]
[tex]= 499444.44[/tex] V/m
Now,
→ [tex]E_2 = \frac{8.99\times 10^9\times 4.0\times 10^{-6 }}{(20\times 10^{-2})^2}[/tex]
[tex]= 89900 0[/tex] V/m
hence,
The net electric field (E) be:
= E₁ + E₂
= [tex]499444.44+899000[/tex]
= [tex]1398444.4 4[/tex] or, [tex]1.39\times 10^6[/tex] V/m
Thus the above answer is correct.
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