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(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge...

Question:

(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below.

(b) If a charge of {eq}-3.80 \space \mu C {/eq} is placed at this point, what are the magnitude and direction of the force on it?

Electric Field

The electric field is qualitatively defined as the electrostatic force per unit charge when placed near a charged particle. The electric field due to point charge particle at a distance from the center of the particle is given by

{eq}\begin{align} E = \frac{kq}{r^2} \end{align} {/eq}

Where q is the charge on the particle and r is the distance from the center of the charge.

Answer and Explanation:

Data Given

Let us redraw the diagram showing the electric field at point P due to all charges,

Electric Field at P

Now the net electric field at point P is given by

{eq}\begin{align} E = E_1 - E_2 + E_3 \end{align} {/eq}

{eq}\begin{align} E = \frac{kq_1}{AP^2} - \frac{kq_2}{BP^2} + \frac{kq_3}{CP^2} \end{align} {/eq}

{eq}\begin{align} E = \frac{9 \times 10^9 \ \rm Nm^2C^{-2} \times 6 \times 10^{-6} \ \rm C}{(0.02 \ \rm m)^2} - \frac{9 \times 10^9 \ \rm Nm^2C^{-2} \times 1.50 \times 10^{-6} \ \rm C}{(0.01 \ \rm m)^2} + \frac{9 \times 10^9 \ \rm Nm^2C^{-2} \times 2.0 \times 10^{-6} \ \rm C}{(0.03 \ \rm m)^2} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ \ E = 2.0 \times 10^7 \ \rm N/C \ }} \end{align} {/eq}


Part B)

  • Charge placed at point P {eq}q = -3.80 \times 10^{-6} \ \rm C {/eq}

The force on the given charge is

{eq}\begin{align} F = qE \\ F = -3.80 \times 10^{-6} \ \rm C \times 2.0 \times 10^{7} \ \rm N/C \\ F = -76 \ \rm N \end{align} {/eq}

Now magnitude of the the force

{eq}\begin{align} \color{blue}{\boxed{ \ F = 76 \ \rm N \ }} \end{align} {/eq} and direction is towards the left.


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