Question

An electron, placed in an electric field, experiences a force F of $1\, N.$ What are the magnitude and direction of the electric field E at the point where the electron is located $(e = 1.6 \times 10^{-19} C)$ ?

• A. $\frac {1}{e} N/C, F$ and E are along the same direction
• B. $\frac {1}{e} N/C, F$ and E are against each other
• C. $\frac {1}{e} N/C, F$ and E are perpendicular
• D. $N/C, F$ and E are against each other

Answer 1

Answer: (B)

$\frac {1}{e} N/C, F$ and E are against each other

Answer 2

Magnitude of electric field is given as,
$E=\frac{F}{q}$
where, $F$ is the magnitude of the electrostatic force and $q$ is the charge
Here, $F=1\, N$
and $q=e=1.6 \times 10^{-19} C$
$\Rightarrow E=\frac{1}{e} N / C$
Force experienced by a negative charge is in the opposite direction to the electric field. As, for a negatively charged particle, its electric field vector at each point is directed radially inwards.
Thus, the magnitude and direction of $E$ is $\frac{1}{e} N / C$ and $E$ and $F$ are against each other.