Question

The electric field intensity just sufficient to balance the earth’s gravitational attraction of an electron will be:
(Given mass and charge of an electron respectively are $9.1 \times {10^{ - 31}}kg$ and $1.6 \times {10^{ - 19}}C$ )

Answer 1

Hint: The force on the electron exerted by the field is qE and the gravitational force acting on the electron is mg.
We know the two forces acting on the electron are that exerted by the field and the gravitational force. So, we will equate the forces and find out the electric field intensity.

Complete step by step answer:
We will first equate the forces acting on the electron that is,
 \[qE = mg\]
Where q = charge of electron
E= electric field
m=mass of electron which is equal to $9.1 \times {10^{ - 31}}kg$
g=acceleration due to gravity which is equal to $10$
So, substituting the given values in the above equation,
 \[qE = mg\]
 $ - eE = mg$
as charge on electron is equal to –e
 $E = - \dfrac{{9.1 \times {{10}^{ - 31}} \times 10}}{{1.6 \times {{10}^{ - 19}}}} = - 5.6 \times {10^{ - 11}}\frac{N}{C}$
Therefore, the electric field required to balance the earth’s gravitational attraction on an electron is
 $ - 5.6 \times {10^{ - 11}}\frac{N}{C}$

Hence the correct solution is option A

Additional information:
Both electric and gravitational fields are similar in many ways such as they both involve magnitude and direction and they impart forces between objects.

Note:
The direction of force on an electron will be in the opposite direction to the electric field as the electron is a negatively charged particle so we should be careful while assigning the direction of attractive forces on the electron. The direction of gravitational force is a vector pointing downwards. If the question involved proton force it would be in the same direction as the electric field.