Question

Two metal plates A and B are connected to cell of emf 2V is shown in the figure

1. Redraw the diagram and draw lines of forces to represent electric field.

2. Calculate the value of electric field between A and B.

3. A charged particle starting from rest moves in the opposite direction of electric field, is it an electron or proton?

4. Calculate acceleration of above particle.

Answer 1

1. Diagram and draw lines of forces to represent electric field:

2. E = \(\frac{v}{d}\) = \(\frac {2}{1\times 10^{-3}}\) = 2 x 10³ v/m.

3. Electron. Electron flows from lower potential to higher potential ie. flows from -ve terminal to +ve terminal.

4. F = eE , ma = eE

a = \(\frac{eE}{m}\) = \(\frac{1.6\times10^{-10}\times 2\times10^3}{9.1\times10^{-31}}\)

a = 3.5 × 10¹⁴ m/sec².