the distancde betwen a parallel plate capictor is 5d the positively charged plate is at x=0 and the negatively charged plate is at x=5d. Two slabs of equal thickness d is inserted bet ween them. One is conductor and another is dielectric. the conductor is placed at d and the dielectric is inserted at 3d. Plot the graph of potiential versusdistance graph
Asked by Ankan | 23rd Oct, 2014, 11:39: AM
The potential versus distance graph will be as shown below:
The potential decreases gradually from x = 0 to x = d.
As it encouters a conductor, the potential remains constant throughout the distance. Hence, it remains same from x = d to x = 2d.
Then it decreases again with the same rate as before till x = 3d.
Then, it encounters a dielectric because of which the rate of drop changes.
After this the drop is same as before till x = 5d, i.e., till the negative plate.
Answered by Romal Bhansali | 27th Oct, 2014, 12:17: PM
- 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
- 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.
- 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
- eletric field on the axis of charged ring and draw its diagram
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