Question

(a) What magnitude point charge creates a \({\rm{10,000 N/C}}\) electric field at a distance of \(0.{\bf{250}}{\rm{ }}{\bf{m}}\)? (b) How large is the field at \({\bf{10}}.{\bf{0}}{\rm{ }}{\bf{m}}\)?

Short answer

(a) The magnitude of the point charge is \(6.95 \times {10^{ - 8}}{\rm{ C}}\).

(b) The magnitude of the electric field at \(10.0{\rm{ m}}\) is \(6.25{\rm{ N/C}}\).

Step-by-step solution

Electric field

Electric field is a vector quantity, is a space around the charge in which the other charge experiences some electrostatic force.

The expression for the electric field is,

\(E = \frac{{Kq}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant, \(q\) is the magnitude of the charge, and \(r\) is the distance.

The magnitude of the charge

(a)

The magnitude of the charge can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the magnitude of charge.

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

Here, \(E\) is the magnitude of the electric field \(E = 10000{\rm{ N/C}}\), \(r\) is the distance of the point from the charge \(\left( {r = 0.250{\rm{ m}}} \right)\), and \(K\) is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\).

Substituting all known values,

\(\begin{aligned} q &= \frac{{\left( {10000{\rm{ N/C}}} \right) \times {{\left( {0.250{\rm{ m}}} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\ &= 6.95 \times {10^{ - 8}}{\rm{ C}}\end{aligned}\)

Hence, the magnitude of the point charge is \(6.95 \times {10^{ - 8}}{\rm{ C}}\).

Magnitude of the electric field

(b)

The magnitude of the electric field is,

\(E = \frac{{Kq}}{{{{r'}^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of charge \(\left( {q = 6.95 \times {{10}^{ - 8}}{\rm{ }}C} \right)\), and \(r'\) is the distance \(\left( {r' = 10.0{\rm{ m}}} \right)\).

Substituting all known values,

\(\begin{aligned} E &= \frac{{\left( {9 \times {{10}^9}{\rm{ N \times }}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {6.95 \times {{10}^{ - 8}}{\rm{ C}}} \right)}}{{{{\left( {10.0{\rm{ m}}} \right)}^2}}}\\ &= 6.25{\rm{ N/C}}\end{aligned}\)

Hence, the magnitude of the electric field at \(10.0{\rm{ m}}\) is \(6.25{\rm{ N/C}}\).