charges of +5µc+10 μc and -10 µc are placed in air at the corners ab and c of an equilateral triangle abc having each side equal to 5 cm. determine the resultant force on the charge at a.
Asked by Sivanigopesh | 26th Apr, 2021, 03:24: PM
Figure shows Forces acting on +5μC charge placed at vertex A.
Force F1 is repulsive force between +5μC charge and +10μC charge.
Force F2 is attractive force between +5μC charge and -10μC charge .
Force between two charges q1 and q2 that are separated by a distance d is given as
Where K = 1/(4πε0 ) = 9 × 109 N m2 C-2 , is Coulomb's constant
Magnniude of each force is same because product of charges is same and distance between charges also same.
Hence Resultant force R = 2 |F| cos60 = ( 2 × 9 × 109 × 50 × 10-12 × cos60) / ( 25 × 10-4 ) = 180 N
Answered by Thiyagarajan K | 26th Apr, 2021, 05:08: PM
- Two point charges 2 C and 3 uC are placed at two comers of an equilateral triangle of side 20 cm in free space. Calculate the magnitude of resultant electric field at the third corner of the triangle. If an & - particle is placed at the third corner, what is the force acting on it? (Charge on & – particle is 3.2 x 10-19 €).
- State Gauss's theorem ? Derive the expression for electric field due Spherical hollow conductor and infinitely charged sheet and linearly charged conductors.
- A simple pendulum is having bob of mass m carrying a charge q and is hanging from ceiling. A uniform electric field E is applied in downward direction then time period of simple pendulum is.
- An electric field E is applied between the plates a and b as shown in the figure a charge particle of mass m and charge q is projected along the direction as shown fig it's velocity v find vertical distance y covered by the partical when goes out of the electric field region
- three charges of 5micro coloumb are placed at the vertices of equilateral triangle of side 10cm the force on 1micro coulomb of charge at centre of triangle will be
- teach me
- In fig three point charges q,-2q are q are placed along the x axis. Show that the electric field at p along the y axis is E =(1/4×3.14×e knot)×(3qa^2/y^4),when y>>a.
- Fig shows a plastic ball of mass m suspended by a light string in a uniform field of intensity e.the ball comes in equilibrium at an angle thita .if ex and ey are the x component and y component of the electric field, find the charge on the ball and the tension in the string
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number