A) \[1.5\text{ }N/C\]
B) \[1.5\,\,\times \,\,{{10}^{-10}}\,N/C\]
C) \[3\,N/C\]
D) \[3\,\,\times \,\,{{10}^{-10}}\,N/C\]
Correct Answer: C
Solution :
The situation is shown in the figure. Plate 1 has surface charge density \[\sigma \] and plate 2 has surface charge density \[-\sigma \]. The electric fields at point P due to two charged plates add up, giving \[E=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{{{\varepsilon }_{0}}}\] Given, \[\sigma \, =\,\,26.4 \times 1{{0}^{-}}^{12}\,C/{{m}^{2}}\] and \[{{\varepsilon }_{0}}=8.85\times {{10}^{-}}^{12}\,\,{{C}^{2}}/N-{{m}^{2}}\] Hence, \[E=\frac{26.4\times {{10}^{-12}}}{8.85\times {{10}^{-12}}}\,\,=\,\,3\,N/C\]
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