Question

two charges (+3uC on the left of the -5uC) are separated by a distance of 5um on a horizontal line. Point P is located 10um above the 3uC charges. a) Draw a sketch of the setting discussed in the question above. Make sure to label all points. b) calculate the field at point P. c) Calculate the potential at point P. d) the force on a +2uC placed at point P​

Answer 1

the concept of vector electric field, the potential electric as a scalar we can find the total electric field, the electric potential and the force on a charge at a given point, the answers are

b) E = (1.61 i ^ + 5.92 j ^) 10¹⁴ N/C

   E = 6.135 10¹⁴  N /C,   θ = 15.2º

c) V = - 1.3 109 V

d) F = 1.23 10⁹ N,  θ = 15.2º

Given parameters

• The charges q₁ = + 3 10⁻⁶ C and q₂ = -5 10⁻⁶C and q₃ = +2 10⁻⁶ C
• The distance between charges 1 and 2 x = 5 10⁻⁶ m
• The distance between load 1 and the test point y = 10 10⁻⁶ m

to find

   a) system scheme

   b) the electric field at the test point

   c) The electric potential at the test point

   d) The force on q₃ at the test point

a) In the adjoint we can see a diagram of the charges and the distances in the exercise there are also the electric field vectors, the force vectors have the same direction as the electric field vectors.

b) The electric field is a vector magnitude that for point charges is given by

             E = k $\frac{q_i}{r^2}$

Where E is the electric field, k is the Coulomb constant (k=8.99 10⁹N m²/C²), q the charge and r the distance between the charge and the positive test charge

The electric field at test point P is the vector sum of the fields created by each

          E_{total} = E₁ + E₂

          E₁ = k q₁ / y²

          E₂ = k q₂ / r²

bold indicates vectors

let's find the distances  

distance from q₁ to the test point

           r = y = 10 10⁻⁶ m

distance from q₂ to test point, we use Pythagoras' theerema

           r² = x² + y²

           r² = (5² + 10² )10⁻¹²

           r² = 125 10⁻¹² m²

           r = 11.18 10⁻⁶ m

we look for every electric field

           E₁ = 9 10⁹  $\frac{3 \ 10-6 }{(10 \ 10-6)^2 }$

           E₁ = 27 10¹³ N / C

           E₂ = 9 10⁹ 5 10⁻⁶ / 125 10⁻¹²

           E₂ = 3.6 10¹⁴ N / C

One of the easiest way to shorten the total field is by using the sum of the components of each field

             E_{total} = Eₓ i ^ + E_y j ^

              Eₓ = E₂ₓ

              E_y = E₁ + E_{2y}

use trigonometry to find the direction of the E₂ field

           tan θ = y / x

           θ = tan⁻¹ $\frac{y}{x}$

           θ = tan⁻¹ (10/5)

           θ = 63.4º

           cos 63.4 = $\frac{E_{2x}}{E_2}$

           sin 63.4 = $\frac{E_{2y}}{E_2}$

           E₂ₓ = E2 cos 63.4

           E_{2y} = E2 sin 63.4

           E₂ₓ = 3.6 10¹⁴ cos 63.4 = 1.61 10¹⁴ N / C

           E_{2y} = 3.6 10¹⁴ sin 63.4 = 3.22 10¹⁴ N / C

The total field is the sum of the component of the electric field

           Eₓ = E₂ₓ = 1.61 10¹⁴ N / C

           E_y = E₁ + E_{2y} = 2.7 10¹⁴ + 3.22 10¹⁴

            E_y = 5.92 10¹⁴ N / C

 

           E = (1.61 i ^ + 5.92 j ^) 10¹⁴ N / C

c) the electric potential is a scalar quantity given by

            V = k ∑ $\frac{q_i}{r}$

for this case

             V = k ( $\frac{q_1}{y} + \frac{q_2}{r}$)

             V = 9 10⁹ (  $\frac{3}{10} - \frac{5}{11.18 }$ )

             V = 9 10⁹ (0.3 - 0.447)

             V = - 1.3 10⁹ V

d) The electric force is given by the relation

              F = q₀ E

where F is the force, qo the test charge and E the total electric field

Let's find the modulus and direction of the electric field using the Pythagorean theorem

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

             E = $\sqrt{1.61^2 + 5.92^2 }$  10¹⁴

             E = 6.135 10¹⁴ N / c

the direction is found with trigonometry

             tan θ = $\frac{E_y}{E_x}$

             θ = tan⁻¹ $\frac{E_y}{E_x}$

             θ = tan⁻¹ $\frac{1.61}{5.92}$

             θ = 15.2º

the force is

              F = 2 10⁻⁶  6.135 10¹⁴

              F = 1.23 10⁹ N

this force has the same direction of the electric field θ = 15,2º measured counterclockwise from the x axis

In conclusion with the concept of vector electric field, the potential electric as a scalar we can find the total electro field, the electric potential and the force on a charge at a given point, the answers are

b) E = (1.61 i ^ + 5.92 j ^) 10¹⁴

   E = 6.135 10¹⁴ N /C,   θ = 15.2º

c) V = - 1.3 10⁹ V

d) F = 1.23 10⁹ N, θ = 15.2º

learn more about electric field here:  

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