# To find the area of sheet placed in uniform electric field

A charged large metal sheet is plaved into uniform field, perpendicularly to the electric field lines. After placing the sheet into the field, the electeic field on the left side of the sheet is $E1= 5 *10^5$ V/m and on the right it is $E2=3*10^5$ V/m (Both towards the sheet). The sheet experiences a net electric force os $0.08 N$. Find the area of the sheet. Assume external field to remain constant after introducing the large sheet.

I don't understand how this is possible and how do I solve it. I never came across such type of question earlier. Any help would be appreciated.

• Is it like , the sheet contains different charges on both halves ? – Nehal Samee May 13 '18 at 13:30
• Don't know. Not mentioned. – Ice Inkberry May 13 '18 at 13:31
• Will using $E=\frac{\sigma}{\epsilon}$ for both halves give the correct area ? – Nehal Samee May 13 '18 at 13:36
• Ummm sorry for the mistake ... That'll be $E=\frac{\sigma}{2\epsilon}$ if used ... – Nehal Samee May 13 '18 at 13:42
• Ummm, doesn't work. Maybe you could write an answer if you know? – Ice Inkberry May 13 '18 at 15:12

The plate is uniformly charged on both sides with same charge density(as in the case of a thin sheet having uniform charge density containing charges of the same sign).Here lets assume that both parts of the plate are negatively charged. as you can see the electric field which was in one particular direction is changing direction.the plate contains charge of equal and same sign of charges on each side.

Assuming the electric field lines(original that is the external field)is hitting the plate from the left(since you didn't mention it) i.e the external field is from left to right.We know that positive charge has fields away from it and negative charge towards it.Now the contribution of the field by the thin plate comes into play here.

The negative charge is helping the external field by raising its magnitude towards the plate from the left side and the negative charge(on the right side) is changing the direction of the field towards the plate on the right side.So the field produced by the plate is good enough to reverse the direction of the external field.Coming to the question assuming field produced by plate is x and the external field is y magnitude wise.Now on the left side of the plate considering direction of the fields towards the plate-

x+y=5(in the given units and 10 powers)

x-y=3

add both the equations- we get x=4(in given units and 10 powers)

hence field due to the sheet is 'x' caused by the negative charges on it.Now you know the formula

Before that you need to use the force formula on the plate that is net electric field intensity multiplied by the charge(total) on the plate will give you the net force on the plate. Solve for the charge-

net electric field intensity on the plate is=5-3(in their units given)

you know the force now solve for the charge on the plates using the formula of electric field by a thin sheet of charges using the electric field value as 'x' in the formula as the field by the sheet.You should get your answer.