Question

Electric field:

An equilateral triangle is formed from three rods, each of length 10 cm. Two of the rods

carry charge +16 nC uniformly distributed along the length of each rod, while the third carries

charge -16 nC, also uniformly distributed along the rod. What is the magnitude of the electric

field at the center of the triangle?

Answer 1

Let AB and AC be positively charged rods. The field with (magnitude say E) due to AB is directed towards C and that (again equal to E in magnitude) due to AC is directed towards B. The components of these along direction BC cancel each other. the perpendicular component gets added. But the perpendicular component is cos 60 (= (1/2)) times E. So the total electric filed due to rods AB and Ac at the center has magnitude E again but directed towards the mid point, say M of BC. As Rod BC is oppositely charged its field also has magnitude E and directed along the same mid pint M. So the net electric field at the center is 2E and has direction pointed towards the mid point of the negatively charged rod. Now what remains is to find the magnitude E.

E = kq/[a*sq rt{a^2 +(L/2)^2}]. where a is the distance of the point along the perpendicular bisector from the mid-point M. and L is the length of the line of charge.
So in our case
E = {9*(10^9)*10*(10^-9)}/[(0.10*tan 30)*{(0.10*tan 30)^2 + (0.10^2)}^(1/2)] or
= {117/(0.10^2)}*{(cos^2 30)/(sin 30)}= 1.17*(3/2) N/C
So the required electric field = 2E = 1.17*3 = 3.51 N/C towards the mid point of the negatively charge

Answer 2

Answer 3

60° 60° 60°

The centre  of an equilateral triangle is

centre a' = 0.5*(3)^0.5L

here we have length 10cm so perpendicular distance of the rod with centre = 2.88cm

so electric field at point P which is the centre due to rod Ep = { (kQ/a)[a²+(L/2)²]^0.5 }^-1  

Ep due to +16nC = 86653.16 N/C perpendicular and away from rod

Ep due to +16nC = 86653.16 N/C perpendicular and away from rod

Ep due to -16nC = 86653.16 N/C perpendicular and towards rod

the net electric field will be

Enet² = (Ep-Epcos30)² +(2Epcos30)²

so Enet = 150448.58 N/C

Answer 4

Electric field due to a line charge at the centre of the equilateral triangle = $\small \frac{2k\lambda }{z}\left ( \frac{a}{\left ( a^2 +z^2\right )^{1/2}} \right )$ where k = 9 x 10⁹ Nm²C^-2 and $\lambda$ - charge per unit length and z - distance from the line charge and a = half side length.

$\lambda$ ₁= - $\lambda$ ₃since the charges are equal and opposite and z is same for both.

Hence E at centre = $\small \frac{2k\lambda_{2} }{z}\left ( \frac{a}{\left ( a^2 +z^2\right )^{1/2}} \right ) = \frac{2*9*10^{9}*\frac{16}{0.1} *10^{-9}}{\frac{\sqrt{3}*0.1}{6}} \left ( \frac{0.05}{(0.05^{2} + (\frac{\sqrt{3}}{60})^2)^{1/2}} \right )$

= 8.64 x 10⁴ N/C