Answer:
[tex]\frac{k_eQ}{2h}[/tex]
Explanation:
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Given;
Q = total charge
R = radius of cylindrical shell
h = height of cylindrical shell
d = distance of point from the right side of the cylinder
Let the thickness of the cylindrical shell be [tex]dx[/tex] , and the charge [tex]\frac{Qdx}{h}[/tex],
Now, using the formula for finding the electric field due to a ring at a chosen point:
[tex]dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}[/tex]
where [tex]x[/tex] = center of the ring to the point
[tex]k_e[/tex] = electrostatic constant
We integrate on both sides from the limits [tex]d[/tex] to [tex]d + h[/tex] in order to determine the electric field at the point [tex]E[/tex]
[tex]\int\limits dE[/tex] = [tex]\int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}[/tex]
[tex]E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i} = \frac{k_eQ}{2h}[/tex]
The electric field at a point a distance d from the right side of the cylinder is : E = [tex]\frac{K_{e}Q }{2h}[/tex]
Let's' assume thickness of cylindrical shell = dx and charge = [tex]\frac{Qdx}{h}[/tex]
Next step : calculate the electric field due to a ring at a point
dE = [tex]\frac{kex}{(x^2+ R^2)^{\frac{3}{2} } } \frac{Qdx}{h} i[/tex] --- ( 1 )
where : x = centre of the ring to the specific point
ke = electrostatic constant
To determine the electric field constant by integrating equation within limits d to d + h
E = [tex]\int\limits^d_d {\frac{keQdx}{h(x^{2}+R^2)^{\frac{3}{2} } } } \, i = \frac{KeQ}{2h}[/tex]
Hence we can conclude that The electric field at a point a distance d from the right side of the cylinder is : E = [tex]\frac{K_{e}Q }{2h}[/tex].
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