The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

$\Delta V = E d$

where

$\Delta V$ is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

$E=5.00\cdot 10^6 V/m$

$d = 2.50 mm = 2.50\cdot 10^{-3} m$

Substituting, we find

$\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V$

3 0

2 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

<em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0

2 years ago