Question

(5%) Problem 22: In the figure, the point charges are equilateral triangle 28 cm on a side. located at the coners of an qa Ogc Oc theespertfa.com 50% Part (a) Find the magnitude of the electric field, that qa 2.9 nC, qb-5.7 nC, and qe- 1.1 nC. triangular configuration of charges, given newtons per coulomb, at the center of Grade Summary 15% El 1963 Deductions Potential 859% sin() cos() tan() 7 9 HOME Submissions Attempts remaining (Se per attempt) cotan) asin) acos() 4 6 sinh) atan) 2 acotan) detailed view cosh() tanh() cotanh) 0 + END ODegrees Radians VO BACKSPACE CLEAR 5% 3 I give up Submit Feedhack nes dadaction nar fondhaek (5%) Problem 22: In the figure, the point charges are located at the coners of an equilateral triangle 28 cm on a side. Oga Oo O qc Ctheexpertta.com 50% Part (a) Find the magnitude of the electric field, in newtons per coulomb, at the center of the triangular configuration of charges, given that q 2.9 nC.b-5.7 nC. and qe 1.1 nC. E 1963 X Attempts Remain 50% Part (b) Find the direction of the electric field in degrees below the right-pointing horizoatal (the positive x-axis). Grade Summary 5% Deductions e-57.7 95% Potential Submissions 7 HOME tan) cos) sin Attempts remaning (51% per attempt) 4 asin) acos) cotan) einh) oatenO

Answer 1

To determine the electric field at the center , first we need to find the distance of each charge from the center of the Triangle. Since, the given triangle is equilateral, the center will be halfway across the base and 1/3 of the way up the height. Using Phythogores theorem to calculate height.

282 142=24.24cm

So, the distance of each charge to the center is

2 x 24.24cm 3

r = 16.16 cm = 16.16 x 10^-2 m

As we know that the electric field intensity at any point is given by E = KQ/r².

Ea = K

2.9 x 10 Ea =(9 x 1016.16 x 10-2)2

E_{​​​​​​a} = 1000 N/C at a 90⁰ angle below the horizontal.

Similarly,

Eb (9 x 10 -5.7 x 10 (16.16 x 10-2)2

E_{​​​​​​b}​​​ = 1965 N/C at a 30^o angle below the horizontal.

And

Ec (9 x 10 1.1 x 10- (16.16 x 10-2)2

E_{​​​​​C} = 380 N/C at a 30^o above the horizontal.

Now, adding the components by vectors, we get

E= Eqcos(-90)Ecos(-30)Eecos30

E 0(1965 ) (0.8660)(380) (0.8660)

E0(1965) (0.8660)(380) (0.8660) 2030N/C

Similarly, for y component,

Ey Easin(-90) Esin(-30)Eesin30

Ey (1000)(-1)(1965) (-0.5) +(380) (0.5)

So, the net electric field intensity is given by

$E = \sqrt { E_x^2 + E_y^2}$

$E = \sqrt { (2030)^2 + (-1793)^2}$

E = 2708 N/C

Direction of the field is,

9=tanEy ET

$\theta = tan^{-1} \frac {-1793}{2030}$

0-41.45

Or we can say that

$\theta = 41.45^0$ below the horizontal.

Answer are:

Part A) 2708 N/C

Part B) 41.45 below horizontal.