Question

An electron traveling parallel to a uniform electric field increases its speed from 2.0 * 107 m/s to 4.0 * 107 m/s over a distance of 1.2 cm. What is the electric field strength?

Answer 1

Answer:

E = 2.84 x 10⁵ N/ C

Explanation:

given,

initial speed of electron,u = 2 x 10⁷ m/s

final speed of electron, v = 4 x 10⁷ m/s

distance = 1.2 cm

electric field strength = ?

using equation of motion for acceleration of electron

v² = u² + 2 a s

$(4\times 10^7)^2 = (2\times 10^7)^2 + 2 \times 0.012 \times a$

$a = \dfrac{12\times 10^{14}}{2\times 0.012}$

    a = 5 x 10¹⁶ m/s²

now,

Force of electric field is equal to , F = q E

and we also know that F = m a

now,

$E = \dfrac{ma}{q}$

where q is the charge of electron = 1.6 x 10⁻¹⁹ C

m is mass of electron, m = 9.1 x 10⁻³¹ Kg

so,

$E = \dfrac{9.1\times 10^{-31}\times 5\times 10^{16}}{1.6\times 10^{-19}}$

       E = 2.84 x 10⁵ N/ C

the electric field strength is equal to E = 2.84 x 10⁵ N/ C

Answer 2

Answer:

The electric field strength is $2.8\times10^{5}\ N/C$

Explanation:

Given that,

Initial velocity $v_{i}= 2.0\times10^{7}\ m/s$

Final velocity $v_{f}=4.0\times10^{7}\ m/s$

Distance = 1.2 cm

We need to calculate the acceleration

Using equation of motion

$v^2=u^2+2as$

Put the value into the formula

$(4.0\times10^{7})^2=(2.0\times10^{7})^2+2\times a\times1.2\times10^{-2}$

$a=\dfrac{(4.0\times10^{7})^2-(2.0\times10^{7})^2}{2\times1.2\times10^{-2}}$

$a=5\times10^{16}\ m/s^2$

We need to calculate the electric field strength

Using formula of electric force

$F=qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{9.1\times10^{-31}\times5\times10^{16}}{1.6\times10^{-19}}$

$E=2.8\times10^{5}\ N/C$

Hence, The electric field strength is $2.8\times10^{5}\ N/C$