Solveeit Logo
Login

Question

Question: A charged cloud system will create an electric field in the air near the surface of earth. A particl...

A charged cloud system will create an electric field in the air near the surface of earth. A particle of charge −2×10−9C-2\times {{10}^{-9}}C is acted on by a downward electrostatic force of 3×10−6N3\times {{10}^{-6}}N when placed in the field, then what will be the gravitational and electrostatic force, respectively, which has been exerted on a proton placed in this field?
A.1.64×10−26N,2.4×10−16N B.1.64×10−26N,1.5×103N C.1.56×10−18N,2.4×10−16N D.1.5×103N,2.5×10−16N \begin{aligned} & A.1.64\times {{10}^{-26}}N,2.4\times {{10}^{-16}}N \\\ & B.1.64\times {{10}^{-26}}N,1.5\times {{10}^{3}}N \\\ & C.1.56\times {{10}^{-18}}N,2.4\times {{10}^{-16}}N \\\ & D.1.5\times {{10}^{3}}N,2.5\times {{10}^{-16}}N \\\ \end{aligned}

Explanation

Solution

The Coulomb force exerted can be found by taking the product of the charge of the particle and the electric field present there. Using this, the electric field at the point can be found. Then using the same equation the electrostatic force exerted by a proton in this field is to be calculated. This will help you in answering this question.

Complete step-by-step solution
The force exerted can be found by taking the product of the charge of the particle and the electric field present there. Using this, the electric field at the point can be found.
F=qEF=qE
The charge of the particle has been mentioned as,
q=−2×10−9Cq=-2\times {{10}^{-9}}C
The electrostatic force persisting there can be shown as,
F=3×10−6NF=3\times {{10}^{-6}}N
Substituting the values in it will give,
3×10−6=2×10−9×E3\times {{10}^{-6}}=2\times {{10}^{-9}}\times E
Therefore the electric field will be,
3×10−62×10−9=E ⇒E=1.5×103 \begin{aligned} & \dfrac{3\times {{10}^{-6}}}{2\times {{10}^{-9}}}=E \\\ & \Rightarrow E=1.5\times {{10}^{3}} \\\ \end{aligned}
Now the electrostatic force exerted on a proton placed in this field can be found by the same equation as,
F=qEF=qE
The charge of a proton can be mentioned as,
q=1.67×10−19Cq=1.67\times {{10}^{-19}}C
Substituting the values in it will give,
F=1.67×10−19×1.5×103 ∴F=2.505×10−16N \begin{aligned} & F=1.67\times {{10}^{-19}}\times 1.5\times {{10}^{3}} \\\ & \therefore F=2.505\times {{10}^{-16}}N \\\ \end{aligned}
Therefore the electric force and the gravitational force have been calculated. This has been mentioned as option D.

Note: Electric field can be described as the electric force acting per unit charge. The electric field can be taken as a vector quantity having both magnitude and also a direction. The direction of the field can be considered to be the direction of the force it would exert on a positive test charge.