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A solid hemisphere is placed at (0,0,0) having radius r and is uniformly charged Q then find the electric field at the point (d,0,0) assume that d>r. I had tried the question by assuming another half sphere and then the answer that i got was E=kQ/d but the answer was given that the field lies between E and 2E and not exactly E

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    Look up Gauss's law. This is a well known result. – 2017-02-28
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    You should ask this question in Physics.SE. I think this is the motives for this downvotes. – 2017-02-28
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    How does one place a hemisphere at a point? Is this a hemispherical shell? – 2017-02-28
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    Related http://math.stackexchange.com/questions/1148334/volume-integral-of-electric-field-hemisphere-solid?rq=1 – 2017-03-04

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One way to go is the following. The potential $V$ for the electric field $\bf E$ can be calculated as

$$V(x,y,z) = \frac{Q}{4\pi \varepsilon_0}\frac{1}{\sqrt{x^2+y^2+z^2}} + V_0,$$

for some constant $V_0$ and whenever $\sqrt{x^2+y^2+z^2}>r$.

Then the electric field is given by:

$${\bf E}(x,y,z) =- \nabla V(x,y,z) = \frac{Q}{4\pi \varepsilon_0} \left( \frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}}, \frac{z}{(x^2+y^2+z^2)^{3/2}} \right).$$

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    You've written the electric potential and corresponding electric field at an arbitrary point due to a point charge at the origin. The charge is distributed uniformly throughout a hemisphere, not a sphere. – 2017-02-28