1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.
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:
Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes
(6.4)^2 x 10^12
= 40.96 x 10^12 .
Now it's just a matter of mashing out the fraction.
The 'mantissa' (the number part) is
6/40.96 = 0.1465
and the order of magnitude is
10^24 / 10^12 = 10^12 .
Put it all together and you've got
1.465 x 10^11 .
