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Question: The force exerted by an electric field on a charge of 3 micro coulomb is \[6 \times {10^{ - 4}}\] . ...

The force exerted by an electric field on a charge of 3 micro coulomb is 6×10−46 \times {10^{ - 4}} . What is the electric intensity?

Explanation

Solution

Electric field intensity on a unit charge is given by the ratio of force exerted to the charge. Substitute the value of the force exerted and the charge on which it is acting. Divide and calculate the value of electric field intensity.

Complete step-by-step solution
Electric field is defined as the space around an electric charge in which its influence can be felt or measured. Whereas the electric field intensity at a point is the force experienced by a unit positive test charge when placed in vacuum
Electric field intensity in terms of force on unit charge is written as,
E=FqE = \dfrac{F}{q}
Where E is the electric field intensity
F is the force produced between 2 charges
Q is the charge experiencing force
As we know that force exerted by a charge on another charge is:
F=q1q24π∫r2F = \dfrac{{{q_1}{q_2}}}{{4\pi \smallint {r^2}}}
In the question we are given that this force is equal to 6×10−46 \times {10^{ - 4}} , and it is acting of a charge of 3×10−63 \times {10^{ - 6}}
Therefore, the electric field intensity at that point will be
E=6×10−43×10−6E = \dfrac{{6 \times {{10}^{ - 4}}}}{{3 \times {{10}^{ - 6}}}}
E=0.02NCE = 0.02\dfrac{N}{C}

Therefore, the correct answer is E=0.02NCE = 0.02\dfrac{N}{C}

Note: The electric field intensity at any point is defined as the force experienced by the unit positive charge at that point.
E⃗=F⃗q\vec E = \dfrac{{\vec F}}{q}
Where, q→0q \to 0 so that presence of this charge may not affect the source charge and its electric field is not changed, therefore expression for electric field intensity can be better written as
E⃗=Limq→0F⃗q\vec E = \mathop {Lim}\limits_{q \to 0} \dfrac{{\vec F}}{q}