Respuesta :
Answer:
The electric field due to the right ring at a location midway between the two rings is [tex]2.41\times10^{3}\ V/m[/tex]
Explanation:
Given that,
Radius of first ring = 5 cm
Radius of second ring = 20 cm
Charge on the left of the ring = +30 nC
Charge on the right of the ring = -30 nC
We need to calculate the electric field due to the right ring at a location midway between the two rings
Using formula of electric field
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}[/tex]
[tex]E=2.41\times10^{3}\ V/m[/tex]
Hence, The electric field due to the right ring at a location midway between the two rings is [tex]2.41\times10^{3}\ V/m[/tex]
The value of the electric field due to the right ring at a location midway between the two rings is [tex]2.41\times10^3 \rm V/m[/tex].
What is electric field?
Electric field is the property associated with each point in the space when there is charge is present. It can be given as,
[tex]E=\dfrac{qx}{4\pi\varepsilon_0\sqrt[3]{(r^2+R^2)^2} }[/tex]
Given information-
The radius of the big ring is 20 cm.
The radius of the small ring is 5 cm.
The ring on the left carries a uniformly distributed charge of +30 n C.
The ring on the right carries a uniformly distributed charge of −30 n C.
Put the values in the above formula as,
[tex]E=\dfrac{9\times10^9\times30\times10^{-9}\times0.1}{\sqrt[3]{(0.1^2+0.2^2)^2} } \\E=2.41\times10^3 \rm V/m[/tex]
Hence the value of the electric field due to the right ring at a location midway between the two rings is [tex]2.41\times10^3 \rm V/m[/tex].
Learn more about the electric field here;
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