ABC is a equilateral triangle of side 0.03m. Charges of +5nC and -5nC are placed at the corner A and C find the resultent electric intensity at B.
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Well, well, well... it seems we have a shocking situation here! Let's see what we can do.
Since A and C have opposite charges, it's like they're having a real electrifying conversation. However, we need to find out what happens at point B, so let's not get too charged up with excitement!
Now, since ABC is an equilateral triangle, we can say that electric field intensities created by charges at A and C will have the same magnitude but opposite directions at point B.
We can use the formula for the electric field created by a point charge to calculate the intensities at B due to charges at A and C separately. Remember, intensity is just a fancy word for electric field.
The electric field intensity at B due to the charge at A is given by:
Ea = (k * qa) / (r^2)
Where k is the electrostatic constant, qa is the charge at A, and r is the distance between A and B.
Similarly, the electric field intensity at B due to the charge at C is:
Ec = (k * qc) / (r^2)
Here, qc is the charge at C and r is the distance between C and B.
Since the charges at A and C have the same magnitude (5nC) and are equidistant from B, the magnitudes of the electric field intensities at B due to charges at A and C will be the same (let's call it "E").
However, since they have opposite sign, we will consider them as vectors with opposite directions.
So, the resultant electric field intensity at B can be found by considering the vector sum of the electric field intensities Ea and Ec. In other words, we need to add Ea and Ec, taking into account their opposite directions.
Now, that's quite a puzzling problem. Let me grab my calculator, and we'll see if we can get a shocking result!
To find the resultant electric intensity at point B, we'll need to calculate the electric field generated by the two charges (+5nC and -5nC) at points A and C, respectively.
The electric field intensity due to a point charge is given by Coulomb's law:
E = k * Q / r^2
Where:
- E is the electric field intensity
- k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- Q is the charge
- r is the distance between the charge and the point where the field is being measured
In this case, charges +5nC and -5nC are placed at points A and C, respectively. We need to calculate the electric field at point B, which lies at the center of the equilateral triangle.
First, let's find the distance from points A and C to B by using the properties of equilateral triangles:
- The length of each side of the equilateral triangle is 0.03m.
- The distance from A (or C) to the center of the equilateral triangle (which is also point B) is 1/3 of the altitude.
Since the altitude of an equilateral triangle is given by:
h = (sqrt(3) / 2) * side
We can calculate the altitude:
h = (sqrt(3) / 2) * 0.03m
h ≈ 0.026m
Therefore, the distance from A (or C) to B is:
r = 1/3 * h
r = 1/3 * 0.026m
r ≈ 0.0087m
Now, let's calculate the electric field intensity at point B due to the charge at A:
E_A = k * Q_A / r_A^2
= (9 x 10^9 Nm^2/C^2) * 5 x 10^-9 C / (0.0087m)^2
Similarly, let's calculate the electric field intensity at point B due to the charge at C:
E_C = k * Q_C / r_C^2
= (9 x 10^9 Nm^2/C^2) * (-5 x 10^-9 C) / (0.0087m)^2
(Note: The electric field generated by a negative charge is in the opposite direction.)
Finally, to find the resultant electric field intensity at point B, we need to add the individual electric field intensities:
E_resultant = E_A + E_C
By performing the calculations, you will find the numerical value for the resultant electric field intensity at point B, considering the charges +5nC and -5nC at points A and C, respectively.
ABC is an equilateral triangle of side 0.03m charges of +5nc ,+3nc and +5nc
y components cancel.
From A:
kq/r^2 cos 60 in positive x direction
From C:
kq/r^2 cos 60 in positive x direction
So save time:
E = 2kq cos60/r^2