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Byju's Answer

Standard XII

Physics

Stopping Potential

PE = 1/4πϵ0Ze...

Question

$P E = \frac{1}{4 π ϵ_{0}} \frac{Z e . (- e)}{r}$
In this expression should we take $(- e)$ value or $(e)$ value ?

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Solution

We should take (e) value as it is a positive number, on the other hand, the charge of electron is -e.

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Similar questions

Q. If

\bar{E}

denote the complement or negation of an even E, what is the value of P(E) + P(

\bar{E}

) ?

Q. Given that

E

and

F

are events such that

P (E) = 0.6, P (F) = 0.3

and

P (E \cap F) = 0.2

, then the value of

P (E / F)

and

P (F / E)

is:

Q. For any event E,

P (E) + P (¯ ¯¯ ¯ E) = 1

where

¯ ¯¯ ¯ E

stands for 'not E'.

E

and

¯ ¯¯ ¯ E

are called complementary events

If E and F are events such that $P (E) = \frac{1}{4},$
$P (F) = \frac{1}{2}$ and $P (E and F)$ $= \frac{1}{8}$ Find
$(i) P (E or F) (ii) P (not E and not F)$

If E and F are events such that P(E) =, P(F) = and P(E and F) =, find:(i) P(E or F), (ii) P(not E and not F).

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PE = 14πϵ0Ze.(−e)rIn this expression should we take (−e) value or (e) value ?

We should take (e) value as it is a positive number, on the other hand, the charge of electron is -e.

$P E = \frac{1}{4 π ϵ_{0}} \frac{Z e . (- e)}{r}$
In this expression should we take $(- e)$ value or $(e)$ value ?