1. Calculate the magnitude of the electric field at the origin due to the follow

Question

1. Calculate the magnitude of the electric field at the origin due to the following distribution of charges: -q at (x,y) = (a,a), +q at (a,-a), +q at (-a,-a) and -q at (-a,a). Where q = 7.25 × 10-7C and a = 4.95 cm.

2. Set up an 8-point compass at the origin, where north points along the positive y-axis, such as that shown in the diagram to the right. What is the direction of the electric field at the origin created by the charge distribution given in the previous part? (for example, `N',`NE', etc., or if no net field enter `NIL').

Explanation / Answer

here ,

distance of chagres from origin ,

d = sqrt(2) * a

Now, as all the charges are equal and at equal distance

electric field at origin due to all charges will be equal in magnitude

E = k *q/d^2

E = k*q/(2*a^2)

Now, as electric field due to charge at (a , a ) is in same direction as field due to charge at (-a , -a)

and

as electric field due to charge at (-a , a ) is in same direction as field due to charge at (a , -a)

and both these are perpendicular to each other

Enet = sqrt((2 * k*q/(2*a^2))^2 + (2k*q/(2*a^2))^2)

Enet = k*q/(a^2) * sqrt(2)

Enet = 9*10^9 * 7.25 *10^-7/(0.0495^2 ) * sqrt(2)

Enet = 3.765 *10^6 N/C

the electric field at the origin is 3.765 *10^6 N/C

2)

as the electric field points towards the negative charhes

the direction of field is in North direction

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