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

The electric intensity due to a dipole of length 10 cm and having a charge of 500 m C, at a point on the axis at a distance 20 cm from one of the charges in air, is (a) 6.25 × 10 7 N/C (b) 9.28 × 10 7 N/C (c) 13.1 × 10 11 N/C (d) 20.5 × 10 7 N/C



The electric intensity due to a dipole of length 10 cm and having a charge of 500 mC, at a point on the axis at a distance 20 cm from one of the charges in air, is


        (a)    6.25 × 107 N/C


(b)    9.28 × 107 N/C


(c)    13.1 × 1011 N/C


(d)    20.5 × 107 N/C


 


Grade:10

2 Answers

Aditi Chauhan
askIITians Faculty 396 Points
8 years ago

(a)

Given: Length of the dipole (2) = 10cm = 0.1 m or  = 0.05 m

Charge on the dipole (q) = 500 pC = 500 x 10-6 C and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm = 0.25 m.

We know that the electric field intensity due to dipole on the given

Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

Length if dipole = 10cm
Distance from point of axis = 20cm
Charge = 500 μC
The electric field intensity due a dipole at a point on the axial line is given as -
E = 2kpr/(r²-l²)2
where p is the dipole moment of dipole, r is the separation between midpoint of dipole to the observation point.
500 μC on each pole of dipole and separation between two dipoles will be 2l = 10cm
Dipole moment , p = 500 × 100cm
= 500 × 10^-6 C × 0.1 m
= 5 × 10^-5 C.m
Distance between the observation point to one charge, d = 20cm
Distance between observation points to midpoint of dipole moment , r = (d + x/2) = 20cm + 5cm = 25cm = 0.25m
E = 2 × 9 × 10^9 × 5 × 10^-5 × 0.25/(0.25² - 0.05²)²
= 90 × 10⁴ × 0.25/(0.30)²(0.20)²
= 2.25 × 10^5/(0.09 × 0.04)
= 2.25/(36) × 10^9
= 225/36 × 10^7
= 6.25 × 10^7 N/C
Thus, one of the charges in air, is 6.25 × 10^7 N/C.

Thanks and Regards

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