Question

an alpha particle placed in a
electric field experiences a force of 6.4×10^-14 N . what is the electric field ?
​

Answer 1

the electric field is given by  E =F/q

charge on alpha particle is = 2e

                                              =2*1.6*10^-19

                                              = 3.2*10^-19 C

therefore, the electric field = 6.4*10^-14/3.2*10^-19

                                               = 2*10^5 V/m

Answer 2

Answer:

The electric field is given = $2*10^{5} V/m$

Explanation:

• Alpha particles are attracted to the negatively charged plate. This confirms that they must be positively charged as unlike charges attract. Alpha particles are helium nuclei; they contain 2 protons which gives them their positive charge.
• Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field (E) at any given location in this space can be determined by placing a test charge q in the space and measuring the force (F) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge.

        $E =\frac{F}{q}$

• q is the symbol used to represent charge, while n is a positive or negative integer, and e is the electronic charge, $1.60 * 10^{-19}$Coulombs

The electric field is given by$E =\frac{F}{q}$

Charge on alpha particle is = 2e

                                           $=2*1.6*10^{-19}$

                                             $= 3.2*10^{-19} C$

Therefore, the electric field $= \frac{ 6.4*10^{-14}}{3.2*10^{-19}}$                                              

= $2*10^{5} V/m$

Reference Link

• https://brainly.in/question/11838220
• https://brainly.in/question/3180587