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:

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

Answer 2

The electric field strength of the electron is 2.84×10⁵ N/C.

To calculate the electric field strength, we use the formula below.

Formula:

• E = ma/q........................ Equation 1

Where:

• E = Electric field strength of the electron
• m = mass of the electron
• a = acceleration of the electron
• q = charge of an electron.

First, we need to calculate the acceleration of the electron.

• v² = u²+2as................ Equation 2

Where:

• v = final velocity of the electron
• u = Initial velocity of the electron
• a = acceleration of the electron
• s = distance covered by the electron.

From the question,

Given:

• v = 4.0×10⁷ m/s
• u = 2.0×10⁷ m/s
• s = 1.2 cm = 1.2×10⁻² m.

Substitute these values into equation 2

• (4.0×10⁷)² = (2.0×10⁷)²+2(1.2×10⁻²)a

Solve for a

• 16×10¹⁴ = 4×10¹⁴+2.4×10⁻²×a
• a = (16×10¹⁴)-(4×10¹⁴)/(2.4×10⁻²)
• a = (12×10¹⁴)/(2.4×10⁻²)
• a = 5×10¹⁶ m/s²

Also, given the following constant,

• m = 9.1×10⁻³¹ kg
• q = 1.6×10⁻¹⁹ C.

Substitute these values into equation 1

• E = (5×10¹⁶)(9.1×10⁻³¹)/(1.6×10⁻¹⁹)
• E = 28.44×10⁴
• E = 2.84×10⁵ N/C

Hence, the electric field strength of the electron is 2.84×10⁵ N/C.

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