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Question: An alpha particle is kept in an electric field of \(15 \times 10^{4} NC^{-1}\) . Calculate the force...

An alpha particle is kept in an electric field of 15×104NC−115 \times 10^{4} NC^{-1} . Calculate the force on the particle. An electron is separated from a proton through a distance of 0.53A0.53 A. Calculate electric field at the location of electrons​.

Explanation

Solution

We have an alpha particle having charge equal to twice of that electron. It is placed in an electric field and then electric force is the product of charge and electric field. Distance between electron and proton is given, we can find the electric field due to proton at the location of the electron.

Complete step-by-step solution:
Given: Electric field, E=15×104NC−1E=15 \times 10^{4} NC^{-1}
Charge on alpha particle, q=2e=2×1.6×10−19=3.2×10−19Cq = 2e =2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} C
We have to find the force on a particle. It is given by –
F=qEF=qE
  ⟹  F=3.2×10−19×15×104\implies F =3.2 \times 10^{-19} \times 15 \times 10^{4}
  ⟹  F=4.8×10−14\implies F = 4.8 \times 10^{-14} N
Distance between electron and proton, r is 0.53A0.53 A.
Electric field is given by-
E=Kqer2E=\dfrac{K q_{e}}{r^{2}}
Where, qe q_{e} is the electron charge.
qe=1.6×10−19q_{e} = 1.6 \times 10^{-19}
r=0.53×10−10mr = 0.53 \times 10^{-10} m
K=9×109Nm2C−2K=9 \times 10^{9} N m^{2} C^{-2}
After putting all values, we get-
E=9×109×1.6×10−19(0.53×10−10)2E=\dfrac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{\left(0.53 \times 10^{-10} \right)^{2}}
  ⟹  E=5.12×1011NC−1\implies E = 5.12 \times 10^{11} N C^{-1}
Hence, the electric force on the particle is 4.8×10−144.8 \times 10^{-14} N and the electric field at the location of the electron​ is 5.12×1011NC−1 5.12 \times 10^{11} N C^{-1}.

Additional Information: - Lines of force begin from a positive charge and end to a negative charge and Electric lines of force are always perpendicular to the surface of the charged body. Two electric lines of force cannot meet each other. There are no lines of force inside the conductor. The tangent to the line of force indicates the direction of the electric field and electric force.

Note: The significance of repulsion or attraction within charged objects is the electric force. Electric field due to a point charge is pointed away from charge if it is positive, and if it is negative then pointed towards the charge, and its magnitude decreases with the square of inverse the distance from the charge.