Question

(I) Determine the magnitude and direction of the electric field 21.7 cm directly above an isolated charge\({\bf{33}}{\bf{.0}} \times {\bf{1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ C}}\).

Short answer

Themagnitude of the electric field is\(6.3 \times {10^6}{\rm{ N/C}}\). The direction of the electric field is upward.

Step-by-step solution

Understanding the electric field on an isolated charge

An electric field due to an isolated charge is the amount of force acting on an isolated charge situated at a specific point.It is inversely proportional to the distance square and directly proportional to the charge on an isolated charge.

Identification of the given data

The given data can be listed as

• The distance between the electric field and the isolated charge is\(r = 21.7{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 0.217{\rm{ m}}\).
• The charge on an isolated charge is\(q = 33 \times {10^{ - 6}}{\rm{ C}}\).
• The Coulomb’s law constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

Determination of the magnitude and the direction of an electric field

The magnitude of an electric field can be expressed as

\(E = k\frac{q}{{{r^2}}}\).

Substituting the values in the above equation,

\(\begin{aligned}{c}E = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2} \times \left\{ {\frac{{33 \times {{10}^{ - 6}}{\rm{ C}}}}{{{{\left( {0.217{\rm{ m}}} \right)}^2}}}} \right\}\\ = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2} \times 7 \times {10^{ - 4}}{\rm{ C/}}{{\rm{m}}^2}\\ = 6.3 \times {10^6}{\rm{ N/C}}\end{aligned}\).

Thus, the magnitude of the electric field is\(6.3 \times {10^6}{\rm{ N/C}}\).

The direction of an electric field is away from an isolated point charge. The electric field is above an isolated charge. So, it is acting in an upward direction, which is directed away from an isolated charge.

Thus, the direction of the electric field is upward.