A 2 V battery is connected to two parallel plates a distance 55 cm apart. If a #-4.30 xx 10^-4 C# charge was placed in the electric field, what force would it feel?

Question

What work is needed to move the #-4.30 xx 10^-4 C# charge from 0.138 m from the ground plate to 0.413 m from the ground plate?

Amelia Adams · Accepted Answer

Electric field is given by  #vecE=V/dhat r#
direction of #hat r# is from positive plate to negative plate. Inserting given values we get #|vecE|=2/0.55Vm^-1# .....(1) Force experienced by a charged particle #q# in an electric field is #vecF=qvecE# Hence , using (1) we get #vecF=2/0.55xx(-4.30xx1^-4)hat r#
#|vecF|=1.56xx10^-3N#
Direction is opposite to that of #hat r#, i.e., towards positive plate Work done in moving the charge is #W=vecFcdot vecs# As the angle between displacement and force vectors is #0^@#,
#costheta# in the dot product #=cos0^@=1# #:.W=|vecF| |vecs|#
#=>W=1.56xx10^-3xx (0.413-0.138)#
#=>W=1.56xx10^-3xx (0.413-0.138)#
#=>W=4.3xx10^-4J#

Lily Adams · Answer

undefined The force experienced by a charge in an electric field can be calculated using the formula:

$$ F = \frac{q \cdot E}{d} $$

Where:
- $ F $ is the force experienced by the charge (in Newtons)
- $ q $ is the magnitude of the charge (in Coulombs)
- $ E $ is the strength of the electric field (in Newtons per Coulomb)
- $ d $ is the distance between the parallel plates (in meters)

Given:
- $ q = -4.30 \times 10^{-4} $ C
- $ d = 55 $ cm $ = 0.55 $ m

The strength of the electric field ($ E $) can be calculated using the formula:

$$ E = \frac{V}{d} $$

Where:
- $ V $ is the voltage (in volts)

Given:
- $ V = 2 $ V
- $ d = 0.55 $ m

Substituting the given values into the formula:

$$ E = \frac{2}{0.55} = 3.64 \, \text{N/C} $$

Now, substituting $ q $, $ E $, and $ d $ into the first formula:

$$ F = \frac{-4.30 \times 10^{-4} \cdot 3.64}{0.55} $$

$$ F = -2.82 \times 10^{-3} \, \text{N} $$

Therefore, the force experienced by the charge is $ -2.82 \times 10^{-3} $ Newtons.