Answer:
Electric field, [tex]E=1.80\times 10^7\ N/C[/tex]
Explanation:
It is given that,
Charge on the ring, [tex]Q=7.36\mu C=7.36\times 10^{-6}\ C[/tex]
Radius of ring, r = 3.33 cm = 0.0333 m
Distance from axis of ring, x = 4.2 cm = 0.042 m
The electric field at the axis of ring is given by :
[tex]E=\dfrac{kxQ}{(x^2+r^2)^{3/2}}[/tex]
[tex]E=\dfrac{9\times 10^9\times 0.042\times 7.36\times 10^{-6}}{(0.042 ^2+0.0333 ^2)^{3/2}}[/tex]
[tex]E=1.80\times 10^7\ N/C[/tex]
So, the magnitude of electric field is [tex]1.80\times 10^7\ N/C[/tex] . Hence, this is the required solution.
Answer:
The magnitude of the electric field is [tex]17.77\times10^{6}\ N/C[/tex]
Explanation:
Given that,
Charge [tex]Q = 7.36\ \mu C[/tex]
radius r = 3.33 cm
Distance a = 4.20 cm
We need to calculate the distance d
According to figure
[tex]d=\sqrt{r^2+a^2}[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\int{dE\cos\theta}[/tex]
[tex]E=\int{\dfrac{k dq}{d^2}\cos\theta}[/tex]
[tex]E=\dfrac{k}{d^2}\cos\theta\int{dq}[/tex]
[tex]E=\dfrac{kQ}{d^2}\times\dfrac{a}{d}[/tex]
[tex]E=\dfrac{kQa}{d^3}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times7.36\times10^{-6}\times4.20\times10^{-2}}{(5.39\times10^{-2})^3}[/tex]
[tex]E=17.77\times10^{6}\ N/C[/tex]
Hence, The magnitude of the electric field is [tex]17.77\times10^{6}\ N/C[/tex]
