A dipole system comprises of two charges q_1 = -15 \space nC located at (x = -5 \space cm, 0) and...

Question:

A dipole system comprises of two charges {eq}q_1 = -15 \space nC {/eq} located at {eq}(x = -5 \space cm, 0) {/eq} and {eq}q_2 = +15 \space nC {/eq} located at {eq}(x = +5 \space cm, 0) {/eq}.

a. Determine the direction and magnitude of electric field at points {eq}P_1 (x = 0, y = 0) {/eq} and {eq}P_3 (x = +9 \space cm, 0) {/eq}.

b. Determine the potential at points {eq}P_1 {/eq} and {eq}P_2 (0, +5 \space cm). {/eq}

c. If we place a charge {eq}q = +5 \space nC {/eq} at {eq}P_1 {/eq}, calculate the magnitude and direction of total electric force on {eq}q {/eq}.

Electric Field and Electric Force:

Electric field due to a point charge is given by the Coulomb's Law: {eq}\vec E = \dfrac {kq}{r^2}\hat r {/eq}, where {eq}k = 9\times 10^9 \ N\cdot m^2/C^2 {/eq} is the Coulomb's constant, q is the charge value, and r is the distance to the point of observation. The direction of electric field coincides with the force direction acting on the unit positive charge placed at the point of observation. The electric potential is also given by Coulomb's Law:

{eq}\varphi = \dfrac {kq}{r} {/eq}. The field and potential due to several charges is the vector or scalar sum of fields or potentials of individual charges, respectively.

Answer and Explanation:

a) Electric field at the point {eq}P_1 (0 \ cm, 0 \ cm) {/eq} is given by:

{eq}\vec E_1 = \left [\dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times (-1.5\times 10^{-8} \ C)}{(0.05 \ m)^2} - \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times 1.5\times 10^{-8} \ C}{(0.05 \ m)^2} \right ] \hat x = \boxed{\color{blue}{(-1.08\times 10^5 \ N/C)\hat x}} {/eq}

Electric field at the point {eq}P_3 (9 \ cm, 0 \ cm) {/eq} is determined as follows:

{eq}\vec E_3 = \left [\dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times (-1.5\times 10^{-8} \ C)}{(0.09 \ m- (-0.05 \ m) )^2} + \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times 1.5\times 10^{-8} \ C}{(0.09 \ m - 0.05 \ m)^2} \right ] \hat x \approx \boxed{\color{green}{(7.75 \times 10^4 \ N/C)\hat x}} {/eq}


b) Electric potential at the point {eq}P_2 (0 \ cm , 5 \ cm) {/eq} is given by the following expression:

{eq}\varphi_2 = \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times (-1.5 \times 10^{-8} \ C)}{\sqrt {(-0.05 \ m)^2 + (0.05 \ m)^2}} + \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times 1.5 \times 10^{-8} \ C}{\sqrt {(0.05 \ m)^2 + (0.05 \ m)^2}} = \boxed {\color{orange}{0 \ V} } {/eq}

Electric potential at the point {eq}P_1 (0 \ cm , 0 \ cm) {/eq} is given by the following expression:

{eq}\varphi_1 = \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times (-1.5 \times 10^{-8} \ C)}{0.05 \ m} + \dfrac {9\times 10^9 \ N\cdot m^2/C^2 \times 1.5 \times 10^{-8} \ C}{0.05 \ m} = \boxed {\color{cyan}{0 \ V} } {/eq}


c) The total force on the charge {eq}q = 5 \times 10^{-9} \ C {/eq} at the point {eq}P_1 (0 \ cm, 0 \ cm) {/eq} is given by:

{eq}\vec F = q \vec E_1 {/eq}

Computation yields:

{eq}\vec F = 5 \times 10^{-9} \ C \times (-1.08\times 10^5 \ N/C)\hat x = \boxed {\color{red}{(-5.4\times 10^{-4} \ N)\hat x}} {/eq}


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