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A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point in an electric field. (Indicate the direction with the signs of your answers. Assume that the positive direction is upward.)

(a) What is the electric field (in N/C) at that point? _________ N/C

(b) If a charge q = −1.1 ✕ 10−8 C is placed there, what is the force (in N) on it? _________N

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Answer:

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

Explanation:

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

(a) Given;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

Substitute these values into equation (i) as follows;

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;

Substitute Q = q into equation (i) as follows;

F = q x E

Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

(a) Electric Field = +2×[tex]10^{2}[/tex] N/C

(b) Force = -2.2×[tex]10^{-6}[/tex]N

where + sign indicates upward direction and - sign indicates downward direction

The relation between Force and Electric Field is given by the following equation:

F = q×E

where F is the force acting on a particle having a charge q under an electric field E.

(a) Given, q = 2.3×[tex]10^{-8}[/tex]C and F = 4.6×[tex]10^{-6}[/tex]N

     to calculate the electric field we use the equation F = q×E

    ⇒ E = F/q

           = 4.6×[tex]10^{-6}[/tex]N ÷ 2.3×[tex]10^{-8}[/tex]C

        E = +2×[tex]10^{2}[/tex] N/C

(b) Given, q = -1.1×[tex]10^{-8}[/tex]C  and we have calculated the electric field E = +2×[tex]10^{2}[/tex] N/C

    F= q×E

      = -1.1×[tex]10^{-8}[/tex]C × +2×[tex]10^{2}[/tex] N/C

    F =  -2.2×[tex]10^{-6}[/tex]N

where + sign indicates upward direction and - sign indicates downward direction

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