Question

Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?

Answer 1

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

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

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

$E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C$

The electric field due to charge 2 is:

$E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C$

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

$E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side$                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

$E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}$

$F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N$

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

Answer 2

(a) The electric field at a point midway between $q_1$ and $q_2$ is obtained to be $2.7\times 10^6 \,N/C$.

(b) The electrostatic force on the third charge $q_3$ situated between $q_1$ and $q_2$ is obtained as 54 N.

The answer can be explained as follows.

Electric Field

Given that the two charges are;

• $q_1 = 50\times 10^{-6}\,C$ and $q_2 = -25\times 10^{-6}\,C$

(a) At the midpoint; $r = 0.5\,m$.

We know that the electric field due to charge $q_1$.

• $E_1 = k\,\frac{q_1}{r^2}$

Where, $k=9\times 10^9\,Nm^2/C$

• $E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C$

The electric field due to charge $q_2$ is given by;

• $E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C$

Therefore, the net electric field in the midpoint is given by;

• $E_{net} =E_2+E_1$
• $\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C$

The direction is towards the right side.

Electrostatic Force

(b) Now, there is another charge $q_3=20\times 10^{-6}$ in the midpoint.

So the force on the charge is ;

• $F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N$

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