Question

Plzzzz help me...... ​

Answer 1

${\mathfrak{\underline{\pink{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{ 4 \: identical \: charges = 2 \mu \: C \:\:\:(each)}$

$\:\:\:\:\bullet\:\:\:\sf{Side \: of \: square = 2 \: m }$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{ Electric \: potential \: at \: centre\:( V_{c})}$

${\mathfrak{\underline{\red{\:\:\: Calculation:-\:\:\:}}}}$

☯ According to the given data,

$\:\:\:\:\bullet\:\:\:\sf{Length \: of \: diagonal= \sqrt{2} a }$

$\:\:\:\:\bullet\:\:\:\sf{ Length \:of \: half \: diagonal = \sqrt{2}}$

☯ As we know that,

$\dashrightarrow\:\: \sf{V_{c} = V_{1} + V_{2} + V_{3} + V_{4} }$

$\sf{(V_{1} = V_{ 2} = V_{3} = V_{4}) }$

$\dashrightarrow\:\: \sf{V_{c} = 4 V_{1} }$

$\dashrightarrow\:\: \sf{ V_{c} = 4 \times \dfrac{k q_{1}}{ r_{1}}}$

$\dashrightarrow\:\: \sf{V_{c} = 4 \times \dfrac{9 \times {10}^{9} \times 2 \times {10}^{ - 6} }{\sqrt{2}} }$

$\dashrightarrow\:\: \sf{V_{c} =4 \times 9 \times \sqrt{2} \times {10}^{9 - 6} }$

$\dashrightarrow\:\: \sf{V_{c} =36\sqrt{2} \times {10}^{3} }$

$\dashrightarrow\:\: \underline{\boxed{\sf{V_{c} =3.6\sqrt{2} \times {10}^{4} \: J /c }}}$

★ The electric potential at center of square is 3.6$\sqrt{2}$ × 10⁴ J/c

Answer 2

Answer:

electic field by the diagonality opposits changes canal each other

E = 0

Voltage:

We know V=

$\frac{kv}{e}$

(E is a vector V is a scalar)

V =

$\frac{ \sqrt{2kv} }{a} \\ \\ total \: v \: = \: \frac{4 \sqrt{2kv} }{a}$

E = 0, V = 0

Step-by-step explanation:

I hope it helps you dear ❣️❣️☺️☺️❤️❤️