Question 10 (1 point) In a right angle, AB2 m, and angle ACB is 63.43 Degree. A

Question

Question 10 (1 point) In a right angle, AB2 m, and angle ACB is 63.43 Degree. A point charge of 2 29 nC is placed at point A and another point charge -3* 29 nC is placed at point C. Calculate the POTENTIAL at B Your Answer Save Question 11 (1 point) 2 charges, 20.15 C each, are located at two vertices B & C of an equilateral triangle ABC with sides 2 cm each. Another charge q is located at point A. Calculate q in micro Coulomb so that net POTENTIAL at the mid point of BC will be ZERO Your Answer Answer Save Question 12 (1 point) Three charges, +34 uC, 34 uC and 34 uC are placed at A (0,5cm), B (5cm,0), C(-5cm,0). Calculate the potential energy of the whole system of charges. Your Answer: Answer units Save Question 13 (0.5 points) Two charges, one is at A with 23.77 nC and other is at B with +9 23.77 nC are seperated by 1 m. Find the distance AC in cm for which electric POTENTIAL at point C is zero. Point C is located on line AB Your Answer Answer units

Explanation / Answer

10)

Potential at B = k*q1/r1 + k*q2/r2

= 9*10^9*(2*29*10^-9/2 - 3*29*10^-9/(2/(tan(63.4 deg))))

= -520.8 V

11)

Potential at mid-point due to two charges at A and B :

2*k*q/r

= 2*9*10^9*(20.15*10^-6)/0.01

= 3.63*10^7 V

So, for net potential to be 0,

k*q'/sqrt(3) = -3.63*10^7

So, 9*10^9*(q')/sqrt(3) = -3.63*10^7

So, q' = 6.98*10^-3 C <------- answer

12)

Vnet = k*q1*q2/r1 + k*q3*q2/r2

= 9*10^9*34*10^-6(34*10^-6/0.1 - 34*10^-6/(0.1*sqrt(2)))

= 30.5 J

13)

k*q1/r = k*q2/(1 - r)

So, k*q/r = k*(9*q)/(1 - r)

So, 1/r = 9/(1 - r)

So, r = 1/10 m = 10 cm from A <-------- answer

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