Answer:
The magnitude and direction of electric field midway between these two charges is [tex]10.8\times10^{5}\ N/C[/tex] along AB.
Explanation:
Given that,
First charge [tex]q_{1}= 20\mu C[/tex]
second charge [tex]q_{2}= 8\mu C[/tex]
Distance = 20 cm
We need to calculate the electric field
For first charge,
Using formula of electric field
[tex]E_{1}= \dfrac{kq_{1}}{r^2}[/tex]
Put the valueinto the formula
[tex]E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}[/tex]
[tex]E_{1}=18\times10^{5}\ N/C[/tex]
Direction of electric field along AB
We need to calculate the electric field
For second charge,
Using formula of electric field
[tex]E_{2}= \dfrac{kq_{2}}{r^2}[/tex]
Put the valueinto the formula
[tex]E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}[/tex]
[tex]E_{2}=7.2\times10^{5}\ N/C[/tex]
Direction of electric field along AO
We need to calculate the net electric field at midpoint
[tex]E_{net}=E_{1}-E_{2}[/tex]
[tex]E_{net}=(18-7.2)\times10^{5}\ N/C[/tex]
[tex]E_{net}=10.8\times10^{5}\ N/C[/tex]
Direction of net electric field along AB
Hence, The magnitude and direction of electric field midway between these two charges is [tex]10.8\times10^{5}\ N/C[/tex] along AB.
