ia totalchange of 100 pC. Ir he radius of the sphere ia so constant om the eleet

Question

ia totalchange of 100 pC. Ir he radius of the sphere ia so constant om the eleetric field from the sarface of the sphere is at a distance of 3.0 (Coulomb c 8.9910 Nm'e * 10* N/C 2.6 x 10" N/C 1.0x 10 NC x 10 N/C E) 3.6x 10 N/C D) 7.4 of 63.0 nC is 10.0 cm from a negative charge of 45.0 nc. The force (Coulomb constant k-8.99-10, S. A positive charge of 63.0 on one of the charges due to the other is approximately A) 1.13 10 N B) 2.54×10-7 N C) 3.02 x 10 N D) 1.25 x 10-N E) 2.54 x 10 N 6. Charges gi and g exert repulsive forces of 15 N on each other. What is the repulsive force when their separation is increased so that their final separation is 140% of their initial separation? A) 1.3 N B) 5.9N C) 9.8N D) 8.7N E) 76N Consider a uniform electric field E (5.0 kN/C) i. What is the flux of this field through a square of side 40 cm if the normal to its plane makes a 45° angle with the x 7. axis? A) 71 N m/C B) 0.14 KN mC C) 0.28 kN-mic

Explanation / Answer

1. Electric field is given by

E= 9*10^9* 100*10^-6/(3.5^2)

E= 7.4*10^4 N/C

----------

2.force between the charges

F= k(q1)(q2)/ r^2

= 9*10^9* 45*63*10^-12/0.1^2

F= 2.54*10^-3 N

---------------

3.

As we know force is inversely proportional to the sqaure of distance between the charges

F2= F1* ( r1/r2)^2

= 15*( 1/1.4)^2

= 7.6 N

-----------------

4. Flux= EA cos x

= 5000* 0.4*0.4* cos 45

= 0.56 KN/c

-------------

Do comment in case any doubt..goodluck

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.