Two large parallel metal plates carry charges +Q and -Q respectively. A test charge q0 placed between them experiences a force F . If the separation between the plates is doubled, then the force on the test charge will be

Question

Two large parallel metal plates carry charges +Q and -Q respectively. A test charge q0 placed between them experiences a force F . If the separation between the plates is doubled, then the force on the test charge will be (A) F (B) 2F (C) F/2 (D) F/4

Tardigrade Admin · Accepted Answer

Initially, the force on charge q0 is <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-fabhbescdtby.png" /> F = q0 E The electric field is, E = (Q/ϵ 0 A) If the separation between the plates is doubled E will remain same as its independent of distance. ∴ F=same

Q. Two large parallel metal plates carry charges $+ Q$ and $- Q$ respectively. A test charge $q_{0}$ placed between them experiences a force $F$ . If the separation between the plates is doubled, then the force on the test charge will be

$F$

$2 F$

$F /2$

$F /4$

Initially, the force on charge $q_{0}$ is

$F = q_{0} E$
The electric field is,
$E = \frac{Q}{ϵ _{0} A}$
If the separation between the plates is doubled $E$ will remain same as its independent of distance.
$∴ F = s am e$

Q. Two large parallel metal plates carry charges +Q and −Q respectively. A test charge q0​ placed between them experiences a force F . If the separation between the plates is doubled, then the force on the test charge will be

F

2F

F/2

F/4

Initially, the force on charge q0​ is F=q0​E The electric field is, E=ϵ0​AQ​ If the separation between the plates is doubled E will remain same as its independent of distance. ∴F=same

Q. Two large parallel metal plates carry charges $+ Q$ and $- Q$ respectively. A test charge $q_{0}$ placed between them experiences a force $F$ . If the separation between the plates is doubled, then the force on the test charge will be

$F$

$2 F$

$F /2$

$F /4$

Initially, the force on charge $q_{0}$ is

$F = q_{0} E$
The electric field is,
$E = \frac{Q}{ϵ _{0} A}$
If the separation between the plates is doubled $E$ will remain same as its independent of distance.
$∴ F = s am e$