Home
Class 12
PHYSICS
Two point charges of +1.5 mu C " and " -...

Two point charges of `+1.5 mu C " and " -1.5 mu C` are placed at the corners A and B of an `-6 mu C` placed at the third corner of the triangle ?

Text Solution

Verified by Experts

The correct Answer is:
90 N, parallel to oppsite side
Promotional Banner

Similar Questions

Explore conceptually related problems

Charges of +5 muC, +10 mu C " and " -10 mu C are placed in air at the corners A,B and C of an equilateral triangle ABC, having each side equal to 5 cm. Determine the resultant force on the charge at A.

Electric charges of + 10 mu C, + 5mu C, - 3mu C and +9 mu C are placed at the corners of a square of side sqrt(2) m the potential at the centre of the square is

Two point charges 2 mu C and 3 mu C are placed at two corners of an equilateral triangle of side 20cm in free space. Calculate the magnitude of resultant electric field at the third corner of the triangle. If an alpha -particle is placed at the third corner, what is the force acting on it? (charge on alpha particle is 3.2 times10^-19 C).

Electric charges of 1mu C, -1 mu C and 2mu C are placed in air at the corners A,B and C respectively of an equilateral triangle ABC having length of each side 10 cm . The resultant force on the charge at C is

Two positive charges of 1mu C and 2 mu C are placed 1 metre apart. The value of electric field in N/C at the middle point of the line joining the charge will be :

Three points charges of 1 C, 2C and 3C are placed at the corners of an equilateral triangle of side 100 cm. Find the work done to move these charges to the corners of a similar equilateral triangle of side 50 cm.

Three point charges of 1C, 2C and 3C are placed at the . corners of an equilateral triangle of side 1m. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side 1.5m.

Two charges of 4 mu C each are placed at the corners A and B of an equilaternal triangle of side length 0.2 m in air. The electric potential at C is [(1)/(4pi epsilon_(0)) = 9 xx 10^(9) (N-m^(2))/(C^(2))]

Two point charges 100 mu C and 5 mu C are placed at points A and B respectively with AB = 40 cm . The work done by external froce in displacing the charge 5 mu C from B to C, where BC = 30 cm , angle ABC = (pi)/(2) and (1)/(4pi epsilon_(0)) = 9 xx 10^(9)Nm^(2)//C^(2)