Question

What is the magnitude of the electric field at a point midway between a −5.6μC and a +5.8μC charge 9.0cm apart? Assume no other charges are nearby. Express your answer using two significant figures.

Answer 1

The magnitude of the electric field at a point midway between two charges can be calculated using the formula E = k * (Q1-Q2) / r^2. Plugging in the values -5.6μC, +5.8μC, and 9.0cm, the magnitude of the electric field is -2.03 x 10^4 N/C.

Step-by-step explanation:

The magnitude of the electric field at a point midway between two charges can be calculated using the formula:

E = k * (Q1-Q2) / r^2

where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

In this case, the charges are -5.6μC and +5.8μC, and the distance is 9.0cm (0.09m).

Plugging these values into the formula:

E = (9 x 10^9 Nm^2/C^2) * ((-5.6μC) - (+5.8μC)) / (0.09m)^2

Simplifying the equation:

E = -2.03 x 10^4 N/C

Answer 2

Magnitude of electric field at mid way is
`5.07* 10^(7) N/C`

Solution:

As per the question:

Q =
`- 5.6\mu m = - 5.6* 10^(- 6) C`

Q' =
`5.8\mu m = 5.8* 10^(- 6) C`

Separation distance, d = 9.0 cm = 0.09 m

The distance between charges at mid-way, O is
`\farc{d}{2} = 0.045 m`

Now, the electric at point O due to charge Q is:

`E = (1)/(4\pi \epsilon_(o)).(Q)/(((d)/(2))^(2))`

`E = (1)/(4\pi \epsilon_(o)).(- 5.6* 10^(- 6))/((0.045^(2))`

`E = (9* 10^(9))(- 5.6* 10^(- 6))/((0.045^(2))`

`E =  (9* 10^(9))(- 5.6* 10^(- 6))/((0.045^(2))`

`E = - 2.49* 10^(7) N/C`

Here, negative sign is indicative of the direction of electric field which is towards the point O

Now, the electric at point O due to charge Q' is:

`E' = (1)/(4\pi \epsilon_(o)).(Q)/(((d)/(2))^(2))`

`E' =  (9* 10^(9))(5.8* 10^(- 6))/((0.045^(2))`

`E' = 2.58* 10^(7) N/C`

Refer to Fig 1.

Since, both the fields are in the same direction:

`E_(net) = E + E' = 2.49* 10^(7) + 2.58* 10^(7) = 5.07* 10^(7) N/C`