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A point charge +q, is placed at a distan...

A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is

A

directed perpendicular to the plane and away from the plane.

B

directed perpendicular to the plane but towards the plane.

C

directed radially away from the point charge.

D

directed radially towards the point charge.

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To solve the problem of finding the electric field at a point P on the other side of an isolated conducting plane, when a point charge +q is placed at a distance d from the plane, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a point charge +q located at a distance d from an isolated conducting plane. - The conducting plane will respond to the electric field created by the point charge. 2. **Induction of Charges on the Conducting Plane**: - The presence of the positive charge +q will induce a negative charge (-q) on the side of the conducting plane that is closest to the charge. - Consequently, a positive charge (+q) will appear on the opposite side of the conducting plane due to the conservation of charge. 3. **Using the Method of Images**: - To analyze the electric field on the side of the conducting plane where point P is located, we can use the method of images. - We replace the conducting plane with an imaginary point charge -q located at a distance d on the opposite side of the plane. 4. **Electric Field Calculation**: - The electric field at point P due to the point charge +q and the imaginary charge -q can be calculated. - The electric field due to a point charge is given by the formula: \[ E = \frac{k \cdot |Q|}{r^2} \] where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge to the point where the field is being calculated. 5. **Direction of the Electric Field**: - The electric field due to the positive charge +q at point P will be directed away from the charge. - The electric field due to the negative charge -q will be directed towards the charge. - Since point P is on the opposite side of the conducting plane, the net electric field at point P will be directed perpendicular to the plane and away from it. 6. **Conclusion**: - The electric field at point P is directed perpendicular to the conducting plane and away from it. ### Final Answer: The electric field at point P on the other side of the conducting plane is directed perpendicular to the plane and away from it.

To solve the problem of finding the electric field at a point P on the other side of an isolated conducting plane, when a point charge +q is placed at a distance d from the plane, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a point charge +q located at a distance d from an isolated conducting plane. - The conducting plane will respond to the electric field created by the point charge. ...
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Knowledge Check

  • A point charge +q is placed at a distance d from an isolated conductin plane. The field at a point P on the other side of the plane is

    A
    directed perpendicular to the plane and away from the plane.
    B
    directed perpendicular to the plane but towards the plane.
    C
    directed radially away from the point charge.
    D
    directed radially towards the point charge.
  • A point charge +q is placed at a distance d from an isolated conducting placen. The field at a point P on the other side of the place is:

    A
    directed perpendicular to the plane and away from the plane.
    B
    directed perpendicular to the plane but towards the plane.
    C
    directed radially away from the point charge
    D
    directed radially towards the point charge.
  • A point charge +q is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of plane is

    A
    directed perpendicular to the plane away from the plane
    B
    directed perpendicular to the plane but towards the plane
    C
    directed radially away from the point charge
    D
    directed radially towards the point charge
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    The method of electrical images is used to solve the electrostatic problems, where charge distribution is not specified completely. The method consists of replacement of given charge distribution by a simplified charge distribution or a signal point charge or a number of point charges [rovided the original boundary conditions are still satisfied. For example consider a system consider a system containing a point charge q placed at a distance d of form an infinite conducting plane. The boundary conditions are : (i) Potential is zero at infinity (ii) Potential is zero at each point on the conducting plane If we replaced the conducting plane by a point charge (-q) placed at a distance 'd' opposite to conducting plane. The charge (-q) is called the electrical image. Now system consists of two charge +q an -q at seperation 2d . If charge +q moves to a distance 'y' from the boundary of conducting plane (now absent), the electrical image -q also moves to the same 'y' from the boundary of conducting plane, so that the new distance between +q and -q is 2y The work done in carrying charge q from distance d to distance y from earthed conducting plane is

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    The method of electrical images is used to solve the electrostatic problems, where charge distribution is not specified completely. The method consists of replacement of given charge distribution by a simplified charge distribution or a signal point charge or a number of point charges [rovided the original boundary conditions are still satisfied. For example consider a system consider a system containing a point charge q placed at a distance d of form an infinite conducting plane. The boundary conditions are : (i) Potential is zero at infinity (ii) Potential is zero at each point on the conducting plane If we replaced the conducting plane by a point charge (-q) placed at a distance 'd' opposite to conducting plane. The charge (-q) is called the electrical image. Now system consists of two charge +q an -q at seperation 2d . If charge +q moves to a distance 'y' from the boundary of conducting plane (now absent), the electrical image -q also moves to the same 'y' from the boundary of conducting plane, so that the new distance between +q and -q is 2y The force between point charge +q and earthed conducting plane is

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