 # An Alpha Particle Is Kept In An Electric Field Of 15×104Nc−1. Calculate The Force Acting On The Particle. An Electron Is Separated From A Proton Through A Distance Of 0.53A. Calculate Electric Field At The Location Of Electrons​.

Given:

α particles have 2 e charge

$$\Rightarrow E = 15 * 10^{4} \\\Rightarrow F = Eq = 15 * 10^{4} * 2 * 1.6 * 10^{-19}\\\Rightarrow 4.8 * 10^{-14} N$$

Therefore, the force acting on it is – $$4.8 * 10^{-14} N$$

Distance between electron and proton, r is 0.53 A.

As we know:

E = $$\frac{Kqe}{r^{2}}$$

Here

K = $$9 * 10^{9} N m^{2} C^{-2}$$

qe = $$1.6 * 10^{-19}$$

r = $$0.53 * 10^{-10} m$$

Now, by substituting the values, we get:

E =$$\frac{9 \;*\; 10^{9} \;* \;1.6 \;* \;10^{-19}}{0.53 \;* \;0.53 \;*\;10^\;{-20}}\\\Rightarrow E = \frac{144 * 10^{10}}{0.2809} \approx \frac{144 * 10^{10}}{28 * 10^{-2}} \\ \Rightarrow E = 5.14 * 10^{12} N/C$$

Therefore, the electric field at the location of electrons is – $$\Rightarrow E = 5.14 * 10^{12} N/C$$

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