# A point charge 50uC is located in the XY plane at the point of position vector ro= 2i + 3j . What is the electric field at the point of position vector r = 8i -5j

Vikas TU
14149 Points
6 years ago
Dear Student,
Distance  = 6i-8j = 10m
Charge = 50 x10^-10
Electric field = kq/r^2
= 9 x10^9 x5 x10^-9 /100
= 0.45 N/C.

Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

Krish Gupta
3 years ago
See you just have to calculate the distance between the two points and then apply the COulomb’s law,
distance vector = r1-r2
Distance  = 6i-8j = 10m
Charge = 50 x10^-10
Electric field = kq/r^2
= 9 x10^9 x5 x10^-9 /100
= 0.45 N/C.
Have fun and enjoy electrostatics !
3 years ago
Dear student,

Let us take vectors,
r0 = 2i + 3j
r1 = 8i -5j
the distance between them is, r = |r1 – r0| = |6i – 8j| = 10m
In vector form, the electric field can be written as,
E = kq(r1 – r0)/r3
= 9 x 10-9 x 5 x 10-6 x (6i – 8j)/103
= 9 x 10-17(3i – 4j) N/C
|E| = 45 x 10-17 N/C

Hope it helps.
Thanks and regards,
Kushagra