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Q18.7-50PE

Expert-verified
Found in: Page 633

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) Find the electric field at the center of the triangular configuration of charges in Figure 18.54, given that $${q_a} = + {\rm{2}}{\rm{.50 nC}}$$, $${q_b} = - {\rm{8}}{\rm{.00 nC}}$$, and $${q_c} = + {\rm{1}}{\rm{.50 nC}}$$. (b) Is there any combination of charges, other than $${q_a} = {q_b} = {q_c}$$, that will produce a zero-strength electric field at the center of the triangular configuration?Figure 18.54 Point charges located at the corners of an equilateral triangle $$25.0{\rm{ cm}}$$ on a side.

(a) The electric field at the center of the equilateral triangle is $$4.36 \times {10^3}{\rm{ N}}/{\rm{C}}$$ and is directed $$34.9^\circ$$ below the positive $$x$$-axis. (b) No, any configuration other than $${q_a} = {q_b} = {q_c}$$ will not produce a zero electric field at the center of the triangular configuration.

See the step by step solution

## Electric field

The electric field is a vector quantity and given as,

$$E = \frac{{Kq}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant, $$q$$ is the magnitude of the charge and $$r$$ is the distance between the charge and the test charge.

For a system of charges, the net electric field at a point is the vector sum of all the electric field at that point.

## (a) Electric field at the center of the triangular configuration

The electric field at the center of the equilateral triangle is represented as,

Electric field at the center of the equilateral triangle

Here, $${E_a}$$ is the magnitude of the electric field due to charge $${q_a}$$, $${E_b}$$ is the magnitude of the electric field due to charge $${q_b}$$, $${E_c}$$ is the magnitude of the electric field due to charge $${q_c}$$, and $$a$$ is the side of the equilateral triangle.

## Distance of the charges from the center of the equilateral triangle

The distance of the charge from the center of the equilateral triangle (length of the centroid) is,

$$r = \frac{a}{{\sqrt 3 }}$$

Here, $$a$$ is the side of the equilateral triangle $$\left( {a = {\rm{25}}{\rm{.0 cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}r = \frac{{{\rm{25}}{\rm{.00 cm}}}}{{\sqrt {\rm{3}} }}\\ = {\rm{14}}{\rm{.4 cm}}\end{array}$$

## Calculating individual electric field

The magnitude of the electric field due to charge $${q_a}$$ is,

$${E_a} = \frac{{K{q_a}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)$$, $${q_a}$$ is the magnitude of the charge $$\left( {{q_a} = {\rm{2}}{\rm{.50 nC}}} \right)$$, and $$r$$ is the distance of the charge from the center of the equilateral triangle $$\left( {r = {\rm{14}}{\rm{.4 cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{E_a} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.50 nC}}} \right)}}{{{{\left( {{\rm{14}}{\rm{.4 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.50 nC}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{ C}}}}{{{\rm{1 nC}}}}} \right)}}{{{{\left[ {\left( {{\rm{14}}{\rm{.4 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{1085 N}}/{\rm{C}}\end{array}$$

The magnitude of the electric field due to charge $${q_b}$$ is,

$${E_b} = \frac{{K{q_b}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)$$, $${q_b}$$ is the magnitude of the charge $$\left( {{q_b} = {\rm{8}}{\rm{.00 nC}}} \right)$$, and $$r$$ is the distance of the charge from the center of the equilateral triangle $$\left( {r = {\rm{14}}{\rm{.4 cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{E_b} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{8}}{\rm{.00 nC}}} \right)}}{{{{\left( {{\rm{14}}{\rm{.4 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{8}}{\rm{.00 nC}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{ C}}}}{{{\rm{1 nC}}}}} \right)}}{{{{\left[ {\left( {{\rm{14}}{\rm{.4 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{3472}}{\rm{.2 N}}/{\rm{C}}\end{array}$$

The magnitude of the electric field due to charge $${q_c}$$ is,

$${E_c} = \frac{{K{q_c}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)$$, $${q_c}$$ is the magnitude of the charge $$\left( {{q_c} = 1.50{\rm{ nC}}} \right)$$, and $$r$$ is the distance of the charge from the center of the equilateral triangle $$\left( {r = {\rm{14}}{\rm{.4 cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{E_c} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.50 nC}}} \right)}}{{{{\left( {{\rm{14}}{\rm{.4 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.50 nC}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{ C}}}}{{{\rm{1 nC}}}}} \right)}}{{{{\left[ {\left( {{\rm{14}}{\rm{.4 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{651 N}}/{\rm{C}}\end{array}$$

## Calculating resultant force

The horizontal component of the electric field is,

$$\begin{array}{c}{E_x} = {E_b}\cos \left( {30^\circ } \right) + {E_c}\cos \left( {30^\circ } \right)\\ = \left( {{E_b} + {E_c}} \right) \times \cos \left( {30^\circ } \right)\end{array}$$

Substituting all known values,

$$\begin{array}{c}{E_x} = \left[ {\left( {{\rm{3472}}{\rm{.2 N}}/{\rm{C}}} \right) + \left( {{\rm{651 N}}/{\rm{C}}} \right)} \right] \times {\rm{cos}}\left( {{\rm{30^\circ }}} \right)\\ = {\rm{3570}}{\rm{.8 N}}/{\rm{C}}\end{array}$$

The vertical component of the electric field is,

$${E_y} = - {E_a} - {E_b}\sin \left( {30^\circ } \right) + {E_c}\sin \left( {30^\circ } \right)$$

Substituting all known values,

$$\begin{array}{c}{E_y} = - \left( {1085{\rm{ N}}/{\rm{C}}} \right) - \left( {3472.2{\rm{ N}}/{\rm{C}}} \right) \times \sin \left( {30^\circ } \right) + \left( {651{\rm{ N}}/{\rm{C}}} \right) \times \sin \left( {30^\circ } \right)\\ = - 2495.6{\rm{ N}}/{\rm{C}}\end{array}$$

The direction of the resultant electric field is,

$$\theta = {\tan ^{ - 1}}\left( {\frac{{{E_y}}}{{{E_x}}}} \right)$$

Substituting all known values,

$$\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{{ - 2495.6{\rm{ N}}/{\rm{C}}}}{{3570.8{\rm{ N}}/{\rm{C}}}}} \right)\\ = - 34.9^\circ \end{array}$$

The magnitude of the resultant field is,

$$E = \sqrt {E_x^2 + E_y^2}$$

Substituting all known values,

$$\begin{array}{c}E = \sqrt {{{\left( {3570.8{\rm{ N}}/{\rm{C}}} \right)}^2} + {{\left( { - 2495.6{\rm{ N}}/{\rm{C}}} \right)}^2}} \\ = 4.36 \times {10^3}{\rm{ N}}/{\rm{C}}\end{array}$$

Hence, the electric field at the center of the equilateral triangle is $$4.36 \times {10^3}{\rm{ N}}/{\rm{C}}$$ and is directed $$34.9^\circ$$ below the positive $$x$$-axis.

## (b) Condition for zero electric field at the center

The electric field at the center will be zero when the following two conditions are satisfied:

• The charges are symmetric (have same magnitude) towards the center so that the force on the test charge is zero.
• The charge on one of them is opposite, so that it balances the force on the test charge.

Hence, any configuration other than $${q_a} = {q_b} = {q_c}$$ will not produce a zero electric field at the center of the triangular configuration.