To find induced charges on the edges of a right pyramid

[Amitkumarr]

Homework Statement

A rigid frame in the shape of a right pyramid is made of conducting rods The base ABCD is a square and the apex O is vertically above the centre of the base. The frame is electrically neutral. When it is placed in a uniform electric field of intensity E pointing from the corner A towards the corner D, total charges induced on the rods DC and OC are known to be q1 and q2 respectively. Now the frame is rotated to make the
electric field pointing from the corner A towards the corner C. What are
the charges induced on each rod?
(refer to the attached figure)

Relevant Equations

Modulus of induced charge on a conductor is directly proportional to the applied electric field intensity.

Due to symmetry of the system,when the frame is rotated to make the electric field point from corner A to corner C,the magnitude of charges induced on these-(AB,BC,CD,DA),(OA,OC),(OB,OD) will be equal(different for each group but same for elements in these groups).
For the sign of induced charges,if we cut a plane OBD,the rods on left hand side will have negative(-) sign and that on the rods lying on right hand side will be having Positive(+)sign.
But,how do I find the magnitude of induced charges on these rods in terms of q1 and q2 ?Should I apply some superposition principle like electric field along the diagonal is superposition of electric field along the edges?If so,how to proceed further?
Also,what about magnitude of induced charges on rods OB and OD?

 

[TSny]

Due to symmetry of the system,when the frame is rotated to make the electric field point from corner A to corner C,the magnitude of charges induced on these-(AB,BC,CD,DA),(OA,OC),(OB,OD) will be equal(different for each group but same for elements in these groups).

Yes

For the sign of induced charges,if we cut a plane OBD,the rods on left hand side will have negative(-) sign and that on the rods lying on right hand side will be having Positive(+)sign.

Yes.

But,how do I find the magnitude of induced charges on these rods in terms of q1 and q2 ?Should I apply some superposition principle like electric field along the diagonal is superposition of electric field along the edges?

Sounds good.

If so,how to proceed further?
Also,what about magnitude of induced charges on rods OB and OD?

Try to find a way to deduce the charge on each of the rods when the field was in the direction from A to D. It might help to think about the case where the field is reversed and points from D to A.

 

[Amitkumarr]

Yes
Yes.

Sounds good.

Try to find a way to deduce the charge on each of the rods when the field was in the direction from A to D. It might help to think about the case where the field is reversed and points from D to A.

I tried but couldn't get the factor by which magnitude of charge will change since in the new situation,few rods are now at π/4 angle to the electric field and earlier some of them were at π/2 angle...please help.

 

[TSny]

Keep in mind that the field is uniform. The field has the same direction and magnitude at all points. So, when the field points in the direction of A to D, how does the charge on OD compare with the charge on OC?

 

[Amitkumarr]

Keep in mind that the field is uniform. The field has the same direction and magnitude at all points. So, when the field points in the direction of A to D, how does the charge on OD compare with the charge on OC?

View attachment 266957

When the electric field is directed from A to D,charge on OD and OC should be equal and of same sign i.e.equal to q2.
So,in the new situation when the electric field points from A to C,if we consider the component of electric field(E) along each rod and use the property that 'Modulus of induced charge on a conductor is directly proportional to the applied electric field' then we get the following charge distribution for the new situation:-
Magnitude of Charge on:1) OD and OB =0(component of E along them is zero)
AB,BC,CD and DA=q1÷√2
OC and OA=√2q2
Thanks for helping.I finally got my answer.
The Top view of the pyramid helped me a lot.

 

[Himansh]

.