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If we put a conductor in a place where there is a uniform electric field, then the field will change. Take for example the case of a conducting cylinder, then the field lines would curve to return something like this:

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But what would happen if the cylinder was held to some potential $V_0>0$ before "turning on" the external electric field? Stated in other words, how would the resulting electric field change if the cylinder had a uniform surface charge density $\sigma$ before being placed inside the external electric field?

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  • $\begingroup$ E resultant can be understand by vector addition of all field,since you mention in your question that v0>0 it mean there is charge whose field is there .hence while doing doing vector addition of field at point where you want the resultant field we also need tk take acciunt of this field $\endgroup$ – Yuvraj Singh... Aug 30 '19 at 6:12
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Because Maxwell’s equations are linear, and we assume linear media, the principle of superposition holds. Thus, the field of a charged cylinder in an external field is simply the field without the added charge plus the field of the charged cylinder.

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  • $\begingroup$ Couldn't that lead to the resulting electric field not being perpendicular to the conductor's surface? Namely, each electric field (the one caused by the charged cylinder and the external one) is normal to the surface, but how can one guarantee that their sum will be normal as well? $\endgroup$ – Tendero Sep 1 '19 at 17:38
  • $\begingroup$ @Tendero Adding two parallel vectors results in a vector parallel to the others. $\endgroup$ – Gilbert Sep 2 '19 at 2:12

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