Question

Each of the protons in a particle beam has a kinetic energy of 2.20 × 10-15 J. protons in a distance of 0.920 m? Answer in units of N/C. What electric field strength will stop these ld

Answer 1

Answer:

The electric field strength is $1.488\times10^{4}\ N/C$

Explanation:

Given that,

Kinetic energy $K.E=2.20\times10^{-15}\ J$

Distance = 0.920 m

We need to calculate the velocity of proton

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$2.20\times10^{-15}=\dfrac{1}{2}\times1.67\times10^{-27}\times v^2$

$v^2=\dfrac{2\times2.20\times10^{-15}}{1.67\times10^{-27}}$

$v=\sqrt{\dfrac{2\times2.20\times10^{-15}}{1.67\times10^{-27}}}$

$v=1.62\times10^{6}\ m/s$

We need to calculate the acceleration

Using equation of motion

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

Put the value into the formula

$a=\dfrac{v^2}{2s}$

$a=\dfrac{(1.62\times10^{6})^2}{2\times0.920}$

$a=1.426\times10^{12}\ m/s^2$

We need to calculate the electric field strength

Using formula of electric field

$F = qE$

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

Put the value in to the formula

$E=\dfrac{1.67\times10^{-27}\times1.426\times10^{12}}{1.6\times10^{-19}}$

$E=1.488\times10^{4}\ N/C$

Hence, The electric field strength is $1.488\times10^{4}\ N/C$