Apne doubts clear karein ab Whatsapp par bhi. Try it now.

CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

40.0 K+ views | 2.0 K+ people like this

Like ShareShare

Answer Text

Solution :

Two charges placed at points A and B are represented in the given figure. O is the mid point of the line joining the two charges. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_014_S01.png" width="80%"> <br> Magnitude of charge located at a, `q_(1) = 1.5 mu C` <br> Magnitude of charge located at B, `q_(2) = 2.5 mu C` <br> Distance between the two charges, `d = 30 cm = 0.3 m` <br> (a) Let `V_(1)` and `E_(1)` are the potential and electric field respectively at O. <br> `V_(1)` = Potential due to charge at A+ Potential due to charge at B <br> `V_(1) = (q_(1))/(4pi in_(0) ((d)/(2)))+(q_(1))/(4pi in_(0)((d)/(2))) = (1)/(4pi in_(0)((d)/(2)))(q_(1)+q_(2))` <br> Where, <br> `in_(0)` = Permittivity of fre space <br> `(1)/(4pi in_(0)) = 9 xx 10^(9) NC^(2)m^(-2)` <br> `:. V_(1) = (9 xx 10^(9) xx 10^(-6))/(((0.30)/(2))) (2.5 + 1.5) = 2.4 xx 10^(5)V` <br> `E_(1)` = Electric field due to `q_(2)` - Electric field due to `q_(1)` <br> `= (q_(1))/(4pi epsilon_(0)((d)/(2))^(2))-(q_(1))/(4pi epsilon_(0)((d)/(2))^(2))` <br> `= (9 xx 10^(9))/((0.30)/(2))^(2) xx 10^(6) xx (2.5 - 1.5)` <br> `= 4 xx 10^(5) V m^(-1)` <br> Therefore, the potential at mid-point is `2.4 xx 10^(5)V` and the electric field at mid-point is `4 xx 10^(5) V m^(-1)`. The field is directed from the larger charge to the smaller charge. <br> (b) consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_014_S02.png" width="80%"> <br> `V_(2)` and `E_(2)` are the electric potential and electric field respectively at Z. <br> It can be observed from the figure distance, <br> `BZ = AZ = sqrt((0.1)^(2)+(0.15)^(2)) = 0.18 m` <br> `V_(2)` = Electric potential due to A + Electric Potential due to B <br> `= (q_(1))/(4pi epsilon_(0)(AZ))+(q_(1))/(4pi epsilon_(0)(BZ))` <br> `= (9 xx 10^(9) xx 10^(-6))/(0.18) (1.5+2.5)` <br> `= 2 xx 10^(5)V` <br> Electric field due to q at Z, <br> `E_(A) = (q_(1))/(4pi epsilon_(0) (AZ)^(2))` <br> `= (9 xx 10^(9) xx 1.5 xx 10^(-6))/((0.18)^(2))` <br> `= 0.416 xx 10^(6) V//m` <br> Electric field due to `q_(2)` at Z, <br> `E_(B) = (q_(2))/(4pi epsilon_(0)(BZ)^(2))` <br> `= (9 xx 10^(9) xx 10^(-6))/((0.18)^(2))` <br> `= 0.69 xx 10^(6) V m^(-1)` <br> The resultant field intensity at Z, <br> `E = sqrt(E_(A)^(2)+E_(B)^(2)+2E_(A)E_(B) cos 2 theta)` <br> Where, `2 theta` the angle `/_ AZ B` <br> From the figure, we obtain <br> `cos theta = (0.10)/(0.18) = (5)/(9) = 0.5556` <br> `theta = cos^([email protected])0.5556 = 56.25` <br> `:. 2 theta =112.5^(@)` <br> `cos 2theta = -0.38` <br> `sqrt((0.416 xx 10^(6))^(2) xx (0.69 xx 10^(6))^(2)+ 2 xx 0.416 xx 0.69 xx 10^(12) xx(0.38))` <br> `= 6.6 xx 10^(5) V m^(-1)` <br> Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is `2.0 xx 10^(5) V` and electric field is `6.6 xx 10^(5)V m^(-1)`.Related Video

FAQs on Electrostatic Potential And Capacitance