Two positive charges of 5.43 µC each are
508 cm apart.
Find the electric field midway between
them.
What is the magnitude of the electric field if
one charge is positive and the other negative,
both of magnitude 5.43 µC?
didn't I see this before?
To find the electric field midway between two positive charges, we can use the formula for the electric field created by a point charge. The formula is given by:
E = k * (q1 / r1^2) + k * (q2 / r2^2)
Where:
- E is the electric field,
- k is the Coulomb's constant (9 x 10^9 N m^2/C^2),
- q1 and q2 are the magnitudes of the charges,
- r1 and r2 are the distances from the charges to the point where we want to find the electric field.
In this case, the two positive charges have magnitudes of 5.43 µC each and are 508 cm apart. Therefore, r1 = r2 = 508 / 2 = 254 cm = 2.54 m. Let's substitute these values into the formula.
E = (9 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2] + (9 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2]
Now let's calculate the result:
E ≈ 5.86 x 10^5 N/C
Therefore, the magnitude of the electric field midway between the two positive charges is approximately 5.86 x 10^5 N/C.
For the second question, when one charge is positive and the other charge is negative, we can find the magnitude of the electric field by using the same formula. However, we need to be careful with the signs of the charges. Since one charge is positive and the other charge is negative, the electric field created by each charge will have opposite directions. Hence, we need to add the magnitudes of the electric fields created by each charge.
Let's substitute the given values into the formula:
E = (9 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (r1)^2] + (9 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (r2)^2]
The result will be the magnitude of the electric field.
Note: To find the direction of the electric field in this case, you need to consider the sign of each charge. The electric field will point away from the positive charge and towards the negative charge.