Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:12

3 Answers

Vikas TU
14149 Points
3 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
askIITians Faculty 82 Points
one year 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 !
Kushagra Madhukar
askIITians Faculty 629 Points
11 months ago
Dear student,
Please find the attached solution to your question.
 
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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free