Nerve cells in your body can be electrically stimulated; a large enough change in a membrane potential triggers a nerve impulse. Certain plants work the same way. A touch to mimosa pudica, the "sensitive plant," causes the leaflets to fold inward and droop. We can trigger this response electrically as well. In one experiment, investigators placed electrodes on the thick tissue at the base of a leaf. The electrodes were 3.5 mm apart. When the electrodes were connected to a 47 μF capacitor charged to 1.5 V, this stimulated a response from the plant.Eventually, all the charge on the capacitor was transferred to the plant. How much charge was transferred?q=What was the approximate electric field between the electrodes?E=

Question

contestada

Respuesta :

skyluke89 skyluke89 · Answer 1 · 2019-05-16T07:57:30+02:00

1. [tex]70.5\cdot 10^{-6} F[/tex]

The relationship between capacitance, charge and voltage across a capacitor is:

[tex]Q=CV[/tex]

where

Q is the charge

C is the capacitance

V is the voltage

In this problem,

[tex]C=47\mu F=47\cdot 10^{-6}F[/tex] is the capacitance

V = 1.5 V

Substituting, we find the charge on the capacitor, that is eventually transferred to the plant:

[tex]Q=(47\cdot 10^{-6}F)(1.5 V)=70.5\cdot 10^{-6} F[/tex]

2. 428.6 V/m

The electric field between the electrodes is given by

[tex]E=\frac{V}{d}[/tex]

where

V = 1.5 V is the potential difference across the electrodes

d = 3.5 mm = 0.0035 m is the distance between the electrodes

Substituting into the equation, we find

[tex]E=\frac{1.5 V}{0.0035 m}=428.6 V/m[/tex]