Can Human Body's Capacitance Destroy Sensitive Electronic Equipment?

  • Thread starter Thread starter Apprentice123
  • Start date Start date
  • Tags Tags
    Capacitance
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
10 replies · 3K views
Apprentice123
Messages
92
Reaction score
0
As a person moves in an environment dry, electric charge accumulates in the body. Since the body is in high voltage, positive or negative, can download by visible sparks or shocks. Consider a human body and separate the soil, with typical capacitance of 150pF.

(a) Load that the body produces a potential of 10,0kV ?

(b) Sensitive electronic equipment can be destroyed by electrostatic discharge of a person. A device can be destroyed by a discharge that releases an energy of 250x10^{-6}J. This corresponds to the voltage on the body ?
 

Discussion

Physics news on Phys.org
negitron said:
Sounds like homework to me. How would YOU go about solving these problems?

I do not understand the problem. What should I calculate?
 
Redbelly98 said:
(a) I am just guessing, but I think "load" means charge. So calculate the charge.

Yes. I also think it load is charge
 
Redbelly98 said:
Have you reviewed the equations that involve capacitance? That is where to start.

Specifically, you want to find the equation which relates capacitance, charge and voltage. And the equation which relates charge, voltage and energy.
 
I think:

a) Q = C.V

Q = (150x10^{-12})x(10x10^{3})
Q = 1.5x10^{-6} coulomb


Alternative (b) I do not understand
 
Apprentice123 said:
I think:

a) Q = C.V

Q = (150x10^{-12})x(10x10^{3})
Q = 1.5x10^{-6} coulomb


Alternative (b) I do not understand

Correct equation and calculation for (a).

Now, for (b), you need an equation that relates capacitance, voltage and the energy stored on that capacitance...
 
berkeman said:
Correct equation and calculation for (a).

Now, for (b), you need an equation that relates capacitance, voltage and the energy stored on that capacitance...

I find
(b) U = 1/2 * C * V^2
V = 1825,74 V