37:Ta39,The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by

\Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}

.32:["$","li","1",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":2}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$37"}}]]}] 38:T7a0,Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by

\Phi_E = E A

, where

A

is the area of the plate.33:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$38"}}]]}] 39:Tcf7,The area of the positive plate is

A = \frac{Q}{\varepsilon_0 V}

, where

Q = C V

is the charge on the positive plate. Therefore, the electric flux through the surface is given by:34:["$","li","3",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":4}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$39"}}]]}] 3a:T1cd1,

\Phi_E = \frac{Q}{\varepsilon_0 V} \frac{V}{d} = \frac{Q}{\varepsilon_0 d} = \frac{C V}{\varepsilon_0 d}

35:["$","li","4",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":5}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3a"}}]]}] 3b:T835,Thus, the answer is

\frac{C V}{\varepsilon_0}

.36:["$","li","5",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":6}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3b"}}]]}]

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