Question

Two charges, Q₁=6.5 nC and Q₂=-6.1x10^-9 C are situated at the diagonal corners A and C of a rectangle of sides AB=DC=19.9 cm and AD=BC=2.4 cm. What is the magnitude of the electric field at point D?

Answer 1

assumption:

let the rectangle is situated along the x and y axis.

let DC be along x-axis and AD be along y -axis.

so for clarity, let the position of all the 4 vertices be:

D: origin (0,0)

C:(19.9,0) cm

B:(19.9,2.4) cm

A: (0,2.4 ) cm

for any charge q , electric field at a distance d is given by

where epsilon=permitivity of free space=8.85*10^(-12)

and the direction of electric field due to a positive charge is away from the charge and due to a negative charge is towards the charge.

step 1:

for Q1:

distance=AD=2.4 cm=2.4*10^(-2) m

charge=Q1=6.5 nC=6.5*10^(-9) C

then using equation for electric field magnitude, we get field magnitude at D due to Q1=1.015625*10^5 N/C

as Q1 is positive , the field will be from A towards D.

hence the field is along -ve y axis.

with vector notation:

field=-1.015625*10^5 j N/C

where j is unit vector along +ve y axis.

step 2:

field due to Q2:

distance=CD=19.9 cm

charge=Q2=-6.1*10^(-9) C

then field magnitude=1386.328 N/C

as the charge is -ve, the field will be from D towards C hence along +ve x axis.

using vector notation, field=1386.328 i N/C

where i is the unit vector along +ve x axis.

hence complete field expression=(1386.328 i - 1.015625*10^5 j)N/C

magnitude of the electric field=sqrt(1386.328^2+(1.015625*10^5 )^2)=1.01572*10^5 N/C