Ahuge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. the upper face carries +95.0 nc/m2 while the lower face carries -25.0 nc/ m2. what is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (ε0 = 8.85 × 10-12 c2/n · m2) a huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. the upper face carries +95.0 nc/m2 while the lower face carries -25.0 nc/ m2. what is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? ( = 8.85 × 10-12 c2/n · m2) 0.00 n/c 6.78 × 103 n/c 7.91 × 103 n/c 1.36 × 104 n/c 3.95 × 103 n/c

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.