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A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict
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The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude: