Question

the conducting plate is charged to a potential of 4,000volt. a second metal plate , charged to a potential of 1,000 volt is brought near the first plate at a distance of 10cm. what is the field intensity at any point between the plates?

Answer 1

Suppose the two plates are A and B.
$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}=4000\mathrm{V} \\ {\mathrm{V}}_{\mathrm{B}}=1000\mathrm{V} \\ \mathrm{So},{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=4000-1000=3000\mathrm{V} \\ \mathrm{Distance},\mathrm{x}=10\mathrm{cm}=0.1\mathrm{m} \\ \mathrm{Field}\mathrm{intensity}, \\ \mathrm{E}=\frac{{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}}{\mathrm{x}} \\ =\frac{3000}{0.1} \\ =3\times {10}^4\mathrm{V}/\mathrm{m} \end{gathered}$$