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lesya [120]
3 years ago
7

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

a point P that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet is E. If the sheet is now stretched so that its sides have length 2d, what is the magnitude of the new electric field at P? The ratio of the new magnitude of E to the initial E is
Physics
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:\frac{1}{4}

Explanation:

Given

Charge Q is spread over non conducting sheet

Electric field due to sheet is E=\frac{\sigma }{2\epsilon _o}

where \sigma =\frac{charge}{area}=\frac{q}{d^2}

E=\frac{q}{2d^2\epsilon _o}

Now if the sheet is stretch to becomes 2d length

E'=\frac{q}{2(2d)^2\epsilon _o}

E'=\frac{1}{4}\times E

E'=\frac{E}{4}

thus the ratio of new and initial electric field is \frac{1}{4}

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