Question

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in an electric field.

a. What is the electric field at that point?
b. If a charge q=1.0 x 10^-8 C is placed there, what is the force on it?

Answer 1

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$, then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$, then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

Answer 2

(a) The electric field of the point charge is 200 N/C.

(b) When the charge changes, the force on the charge in the field is 2 x 10⁻⁶ N.

Electric field of the point charge

The electric field of the point charge in a given is defined as the force per unit of charge.

$E = \frac{F}{q} \\\\E = \frac{4 \times 10^{-6} }{2 \times 10^{-8}} \\\\E = 200 \ N/C$

Force on the charge

When the charge changes, the force on the charge in the field is calculated as follows;

F = Eq

F = (200) x (1 x 10⁻⁸)

F = 2 x 10⁻⁶ N

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