Direction Q.41 to 45 |
Surface charge density is defined as charge per unit da surface area of surface charge distribution i.e., \[\sigma =\frac{dq}{ds}\]. |
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of \[17.0\times {{10}^{-22}}\,C{{m}^{-2}}\]as shown. The intensity of electric field at a point is\[E=\frac{\sigma }{{{\varepsilon }_{0}}}\], where \[{{\varepsilon }_{0}}\]= permittivity of free space. |
Read the above passage carefully and give the answer of the following questions. |
A) \[17\times {{10}^{-22}}N/C\]
B) \[1.5\times {{10}^{-25}}N/C\]
C) \[1.9\times {{10}^{-10}}N/C\]
D) zero
Correct Answer: D
Solution :
(d) zero There are two plates A and B having surface charge densities, \[{{\sigma }_{A}}=17.0\times {{10}^{-22}}C/{{m}^{2}}\] on A and \[{{\sigma }_{B}}=-17.0\times {{10}^{-22}}C/{{m}^{2}}\]on B, respectively. According to Gauss's theorem, if the plates have same surface charge density but having opposite signs, then the electric field in region I is zero. \[{{E}_{1}}={{E}_{A}}+{{E}_{B}}=\frac{\sigma }{2{{\varepsilon }_{0}}}+\left( -\frac{\sigma }{2{{\varepsilon }_{0}}} \right)=0\]You need to login to perform this action.
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