Two charges -1 microcoulumb and 2 microcoulumb are placed at the corner A and B of an equilateral triangle ABC of side 2m. Calculate the electric field at C
To calculate the electric field at point C, we need to determine the individual electric fields due to each charge at that point and then add them together vectorially.
Step 1: Calculate the electric field due to the -1 microcoulumb charge at point C.
The electric field due to a point charge is given by the equation:
E = k * q / r^2
where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance between the charge and the point of interest.
In this case, the charge (q) is -1 microcoulumb (-1 x 10^-6 C), and the distance (r) is the length of one side of the equilateral triangle, which is 2m. Plugging these values into the equation:
E1 = (9 x 10^9 Nm^2/C^2) * (-1 x 10^-6 C) / (2m)^2
Step 2: Calculate the electric field due to the 2 microcoulumb charge at point C.
Similarly, we use the equation:
E = k * q / r^2
where q is 2 microcoulumb (2 x 10^-6 C), and r is the distance between the charge and the point C. Since ABC is an equilateral triangle, the distance r can be found using the Pythagorean theorem. The triangle ABC can be divided in half to form two right-angled triangles, with C as the right angle. The hypotenuse of each right-angled triangle is the distance r we are looking for.
The length of each side of the equilateral triangle is 2m, so the height of the right-angled triangle is equal to the square root of 3 divided by 2 times the length of the side:
Height = (√3 / 2) * (2m) = √3 m
Now, we can use the Pythagorean theorem to find the length of the hypotenuse (r) of the right-angled triangle:
r = √(2m)^2 + (√3 m)^2
r = √(4m^2 + 3m^2)
r = √(7m^2)
r = √7 m
Plugging these values into the equation, we get:
E2 = (9 x 10^9 Nm^2/C^2) * (2 x 10^-6 C) / (√7m)^2
Step 3: Add the two electric fields vectorially.
Since the two electric fields are at 180 degrees to each other (opposite directions), we need to subtract their magnitudes to get the net electric field at point C:
E_net = |E1| - |E2|
where |E1| and |E2| are the magnitudes of E1 and E2, respectively.
Finally, we substitute the calculated electric fields E1 and E2 into the formula to find the net electric field at point C.
A is at (0,0)
B is at (1,sqrt 3)
C is at (2, 0)
qa = -1*10^-6
qb = +2*10^-6
E at C due to A = K (-1*10^-6)/4 i + 0 j
E at C due to B = K [(+2*10^-6)/4] [cos 60 i - sin 60 j ]
so
Ex at C = 10^-6 K [ -1/4 +2cos60/4]
Ey at C = 10^-6 K [-.25 sin 60]
|E| = sqrt(Ex^2 + Ey^2)
tan angle from x = Ey/Ex