Two fixed charges, -4μC and -5μC, are separated by a certain distance. If the charges are separated by 20cm, what is the magnitude of the electric field halfway between the charges?
My textbook says that the formula for electric field is (kq)/r^2.
But Since I have two charges (and both are negative) I don't really know what to put for q.
Would I solve for the electric fields separately and then find their vector sum to get the halfway point electric field?
Or would q be the difference between the two q's?
Please help lead me in the right direction!!! Thank you!
Well, you can find the electric field E due to charge 1 at charge 2, then multiply that field by charge 2 to find the repulsive force (note, same sign charge repels, different signs attract)
HOWEVER, normally you use Coulombs Law
F = k Q1 Q2 / r^2
( Google "Coulomb's Law" )
as you can see, it amounts to the same thing because the Electric field E is in fact the force on a charge of 1 coulomb.
To find the electric field at the halfway point between the two charges, you can first calculate the electric field created by each individual charge and then find the vector sum of the two electric fields.
The formula for the electric field due to a point charge is given by:
E = (k * q) / r^2
Where:
E is the electric field
k is Coulomb's constant (approximately 8.99 x 10^9 Nm^2/C^2)
q is the charge
r is the distance from the charge
Since both charges are negative, it means they will create electric fields that point towards each other.
First, let's calculate the electric field created by the charge of -4μC at the halfway point.
Using the formula, with q = -4μC = -4 x 10^-6 C and r = 10 cm (halfway point between 20 cm distance):
E1 = (k * q) / r^2
= (8.99 x 10^9 Nm^2/C^2 * -4 x 10^-6 C) / (0.1 m)^2
= -3.596 x 10^5 N/C
Similarly, let's calculate the electric field created by the charge of -5μC at the halfway point.
Using the same formula, with q = -5μC = -5 x 10^-6 C and r = 10 cm (halfway point between 20 cm distance):
E2 = (k * q) / r^2
= (8.99 x 10^9 Nm^2/C^2 * -5 x 10^-6 C) / (0.1 m)^2
= -4.495 x 10^5 N/C
Now, to find the total electric field at the halfway point, you can sum the two electric fields together:
E_total = E1 + E2
= -3.596 x 10^5 N/C + (-4.495 x 10^5 N/C)
= -8.091 x 10^5 N/C
Therefore, the magnitude of the electric field at the halfway point between the charges is 8.091 x 10^5 N/C. Note that the negative sign indicates the direction of the electric field.
To find the magnitude of the electric field halfway between the two fixed charges, you indeed need to calculate the electric fields separately for each charge and then find their vector sum. Let's go step by step:
Step 1: Calculate the electric field due to the -4μC charge:
Using the formula for electric field, E = (k * q) / r^2, where k is the Coulomb's constant (9 * 10^9 Nm^2/C^2), q is the charge, and r is the distance, substitute the given values:
E1 = (9 * 10^9 Nm^2/C^2) * (-4 * 10^-6 C) / (0.1 m)^2
E1 = -360 N/C
Step 2: Calculate the electric field due to the -5μC charge:
Using the same formula, substitute the given values:
E2 = (9 * 10^9 Nm^2/C^2) * (-5 * 10^-6 C) / (0.1 m)^2
E2 = -450 N/C
Step 3: Find the vector sum of the electric fields:
Since the electric fields due to both charges have the same direction (because both charges are negative), you can simply add their magnitudes:
E_halfway = |E1| + |E2|
E_halfway = |-360 N/C| + |-450 N/C|
E_halfway = 810 N/C
Therefore, the magnitude of the electric field halfway between the charges is 810 N/C. The negative sign indicates that the direction of the electric field is towards the negative charges.