Answer to Question #137480 in Electric Circuits for mail

Question

2. The plate of a parallel plate capacitor, 5.0 x 10 -3 m apart are maintained at a potential
difference of 5.0 x 10 4 V. calculate the magnitude of the:
(i) Electric field intensity between the plate
(ii) Force on the electron
(iii) Acceleration of the electron
(Electronic charge = -1.6 x 10 -19 C, mass of electron = 9.1 x 10 -31 kg)

Answer 1

1) Since the potential $\phi =\frac W q$ , where $W$ is energy and $q$ is charge, the voltage $U=∆\phi=\frac A q$ , where $A$ is work, a work is equal to $A=q\times E \times ∆r$ , from this follows that

$E=\frac {∆\phi}{∆r}$ $= \frac {U}{∆r}=$ $\frac {5\times 10^4}{5\times 10^{-3}}=10^7$ V.

2) Since $E = \frac {F} {q}$ , where $F$ is the force, $F = E\times q = 10 ^ 7\times 1 .6\times 10 ^ {-19}$ $= 1 .6\times 10 ^ {-12}$ N.

3) According to Newton's first law, $F = m\times a$ , therefore $a =\frac F m =$ $\frac {1 .6\times 10 ^ {-12}} {9 .1\times 10 ^ {-31} }$ $\approx 1 .76\times 10 ^ {18}$ m/s $^ 2$.