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The answer is : perpendicular to the diameter. I don't understand how to arrive at that answer. I'm taking a high school physics course and have covered Electric field and Gauss's law for simple symmetries such as the sphere. Can anyone help me on this?

Edit: The solution says that the field component parallel to the diameter cancels out, which I don't think is correct because the point is away from the centre, and since the hemisphere is uniformly charged so on one side there will be more charge than on the other side. Any thoughts on this?

I have no other method of feedback so I'd appreciate any help I can get.

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  • $\begingroup$ Please explain your problem properly $\endgroup$ – Sourabh Dec 4 '18 at 15:37
  • $\begingroup$ Can you tell me the position properly at which you want to calculate electric field $\endgroup$ – Sourabh Dec 4 '18 at 15:44
  • $\begingroup$ link . Question 1.7 $\endgroup$ – LiqSnk Dec 4 '18 at 15:45
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Let us suppose that there existed a horizontal component too, if we take another identical hemisphere and join the 2 hemispheres to make a perfect sphere, then the vertical components of their individual electric fields would cancel out while the horizontal component would not. This implies that net electric field will exist in the sphere. But as we know field inside a hollow sphere is zero. This imples that horizontal component of field doesnot exists..enter image description here

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  • $\begingroup$ Why would the horizontal components not get cancelled out? $\endgroup$ – LiqSnk Jul 2 at 1:53
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From picture you can conclude for every point charge present on the surface or inwards the hemisphere their exist a corresponding point charge whose electric field's direction cancels the horizontal component of electric field hence only vertical componenet left.

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  • $\begingroup$ But I need the field on a point on the diameter, also away from the centre. You have shown it for a point away from the diameter at the centre $\endgroup$ – LiqSnk Dec 6 '18 at 15:49

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